NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability - Exercise 5.5
Access Exercises of Class 12 Maths Chapter 5 – Continuity and Differentiability
Exercise 5.1 Solutions: 34 Questions (Short Answers)
Exercise 5.2 Solutions: 10 Questions (Short Answers)
Exercise 5.3 Solutions: 15 Questions ( Short Answers)
Exercise 5.4 Solutions: 10 Questions (Short Answers)
Exercise 5.5 Solutions: 18 Questions ( Short Answers)
Exercise 5.6 Solutions: 11 Questions (Short Answers)
Exercise 5.7 Solutions: 17 Questions (Short Answers)
Exercise 5.8 Solutions: 6 Questions (Short Answers)
Miscellaneous Exercise Solutions: 23 Questions (6 Long, 17 Short)
Exercise 5.5
Direction (Q. 1 to 11) : Differentiate the given functions w.r.t. x.
1. cos x cos 2x cos 3x.
Sol. Let y = cos x cos 2x cos 3x,
Taking log on both side, we get
log y = log (cos x cos 2x cos 3x)
[∵ log m × n × l = log m + log n + log l]
or log y = log (cos x) + log (cos 2x) + log (cos 3x)
Differentiating both sides w.r.t. x, we get
$$\frac{d}{dx}\text{log y = }\frac{d}{dx}\text{log(cos x)} +\\\frac{d}{dx}\text{log(\text{cos 2x}) +}\\\frac{d}{dx}\text{log(\text{cos 3x})}\\\frac{1}{y}\frac{dy}{dx} =\frac{1}{\text{cos x}}(-\text{sin x})+\\\frac{1}{\text{cos 2x}}\begin{Bmatrix}-\text{sin 2x}\frac{d}{dx}(2x) +\\\frac{1}{\text{cos 3x}}\begin{Bmatrix}-\text{sin 3x.}\frac{d}{dx}(3x)\end{Bmatrix}\end{Bmatrix}\\\Rarr\space\frac{dy}{dx} =y\lbrace-\text{tan x -2 tan 2x} -\\ 3\text{tan 3x}\rbrace$$
= – y {tan x + 2 tan 2x + 3 tan 3x}
= – cos x cos 2x cos 3x
{tan x + 2 tan 2x + 3 tan 3x}
$$\textbf{2.\space}\sqrt{\frac{\textbf{(x-1)(x-2)}}{\textbf{(x-3)(x-4)(x-5)}}}.\\\textbf{Sol.\space}\text{Let y = }\\\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}$$
Taking log on both sides, we get
$$\text{log y = log}\\\begin{Bmatrix}\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}\end{Bmatrix}^{\frac{1}{2}}\\=\frac{1}{2}\lbrack \text{log}(x-1)(x-2) -\\\text{log}(x-3)(x-4)(x-5)\rbrack\\=\frac{1}{2}\lbrack\text{log}(x-1) + \text{log}(x-2) -\\\text{log}(x-3)-\text{log}(x-4) - \text{log}(x-5)\rbrack$$
Differentiating both sides w.r.t. x, we get
$$\frac{1}{y}\frac{dy}{dx} = \frac{1}{2}\\\begin{bmatrix}\frac{d}{dx}\text{log}(x-1) + \frac{d}{dx}\text{log}(x-2)-\\\frac{d}{dx}\text{log}(x-3)-\\\frac{d}{dx}\text{log}(x-4)-\frac{d}{dx}(x-5)\end{bmatrix}$$
$$=\frac{1}{2}\begin{Bmatrix}\frac{1}{\text{x-1}}(1-0) + \frac{1}{x-2}(1-0)-\\\frac{1}{x-3}(1-0)\frac{1}{x-4}(1-0)-\frac{1}{x-5}(1-0)\end{Bmatrix}$$
$$\Rarr\space\frac{dy}{dx} = \frac{y}{2}\begin{Bmatrix}\bigg(\frac{1}{x-1} + \frac{1}{x-2}\bigg)-\\\bigg(\frac{1}{x-3} + \frac{1}{x-4} + \frac{1}{x-5}\bigg)\end{Bmatrix}\\=\frac{1}{2}\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\\\begin{Bmatrix}\frac{1}{x-1} + \frac{1}{x-2}-\frac{1}{x-3}-\\\frac{1}{x-4}-\frac{1}{x-5}\end{Bmatrix}$$
3. (log x)cos x.
Sol. Let y = (log x)cos x
Taking log on both sides, we get
log y = log {(log x)cos x}
⇒ log y = cos x log (log x)
[∵ log mn = n log m]
Differentiating both sides w.r.t. x, we get
$$\Rarr\space\frac{1}{y}\frac{dy}{dx} =\\\frac{d}{dx}\lbrace\text{cos x log(log x)}\rbrace\\\Rarr\space\frac{1}{y}\frac{dy}{dx} =\text{cos x}\frac{1}{\text{log x}}.\frac{1}{x}+\\\text{log(log x)(- sin x)}\\\begin{pmatrix}\text{Using product rule}\frac{d}{dx}(u.v) =\\u\frac{dv}{dx} + v\frac{du}{dx}\end{pmatrix}\\\Rarr\space\frac{dy}{dx} = y\begin{Bmatrix}\frac{\text{cos x}}{\text{x log x}} -\\\text{sin x log(log x)}\end{Bmatrix}\\=\text{(log x)}^{\text{cos}x}\\\begin{Bmatrix}\frac{\text{cos x}}{\text{x log x}} -\text{sin xlog(log x)}\end{Bmatrix}$$
4. xx – 2sin x.
Sol. Let y = xx – 2sin x
Let u = xx and v = 2sin x
∴ y = u – v
Differentiating both sides w.r.t. x, we get
$$\frac{dy}{dx} = \frac{du}{dx} - \frac{dv}{dx}$$
Now, u = xx ...(i)
Taking log on both sides,
⇒ log u = log xx
⇒ log u = x log x
$$\Rarr\space\frac{1}{u}\frac{du}{dx} = x\frac{d}{dx}\text{log x +}\\\text{log x}\frac{d}{dx}x$$
(Differentiate w.r.t. x)
$$\Rarr\space\frac{1}{u}\frac{du}{dx} = x\frac{d}{dx}\text{log x +}\\\text{log x}\frac{d}{dx}x$$
$$\Rarr\space\frac{1}{u}\frac{du}{dx} =\\x×\frac{1}{x} + \text{log x}\\\Rarr\space\frac{1}{u}\frac{du}{dx} =\text{1 + log x}\\\frac{du}{dx} = x^{x}(\text{1 + log x})$$
v = 2sin x
Taking log on both sides,
⇒ log v = log (2sin x) ⇒ log v = (sin x) log 2
$$\Rarr\space\frac{1}{v}\frac{dv}{dx} =\text{cos x(\text{log 2})}$$
(Differentiate w.r.t. x)
$$\Rarr\space\frac{dv}{dx} = v\lbrack\text{cos x log 2}\rbrack$$
= 2sin x[cos x log 2]
Now, putting the values of
$$\frac{du}{dx}\space\text{and}\space\frac{dv}{dx}\\\text{in Eq. (i)}\\\frac{dy}{dx} = x^{x}[\text{1 + log x}]-\\2^{\text{sin x}}\lbrack\text{cos x log 2}\rbrack$$
5. (x + 3)2(x + 4)3(x + 5)4.
Sol. Let y = (x + 3)2(x + 4)3(x + 5)4
Taking log on both sides, we get
log y = log {(x + 3)2 (x + 4)3(x + 5)4}
or log y = 2 log (x + 3) + 3 log (x + 4) + 4 log (x + 5)
(∵ log (lmn) = log l + log m + log n)
Differentiating both sides w.r.t. x, we get
$$\frac{d}{dx}(\text{log y}) = 2\frac{d}{dx}\text{log}(x+3)+\\3\frac{d}{dx}\text{log}(x+4)+ 4\frac{d}{dx}\text{log(x+5)}\\\Rarr\space\frac{1}{y}\frac{dy}{dx} = 2\bigg(\frac{1}{x+3}\bigg)(1 + 0) +\\3\bigg(\frac{1}{x+4}\bigg)(1 + 0) +\\4\bigg(\frac{1}{x+5}\bigg)(1 + 0)\\\Rarr\space\frac{dy}{dx} = y \begin{Bmatrix}\bigg(\frac{2}{x+3}\bigg) + \bigg(\frac{3}{x+4}\bigg)+\\\bigg(\frac{4}{x+5}\bigg)\end{Bmatrix}$$
$$\Rarr\space\frac{dy}{dx} =(x+3)^{2}(x+4)^{3}(x+5)^{4}\\\begin{Bmatrix}\bigg(\frac{2}{x+3}\bigg) + \bigg(\frac{3}{x+4}\bigg) + \bigg(\frac{4}{x+5}\bigg)\end{Bmatrix}$$
= (x + 3)2(x + 4)3(x + 5)4
$$\begin{Bmatrix}\frac{2(x+4) (x+5) +3 (x+3) (x+5) + 4(x+3)(x+4)}{(x+3)(x+4)(x+5)}\end{Bmatrix}$$
= (x + 3)(x + 4)2(x + 5)3[2(x2 + 9x + 20) + 3(x2 + 8x + 15) + 4(x2 + 7x + 12)]
= (x + 3)(x + 4)2(x + 5)3(9x2 + 70x + 133)
$$\textbf{6.\space}\bigg(\textbf{x +}\frac{\textbf{1}}{\textbf{x}}\bigg)^{\textbf{x}} +x^{\bigg(\textbf{1 +}\frac{\textbf{1}}{\textbf{x}}\bigg)}\textbf{.}\\\textbf{Sol.\space}\text{Let\space y =}\\\bigg(x+\frac{1}{x}\bigg)^{x} +x^{\bigg(1 + \frac{1}{x}\bigg)}\\\text{Again let}\\\space u =\bigg(x +\frac{1}{x}\bigg)^{x},\\ v = x^{\bigg(1 +\frac{1}{x}\bigg)}$$
∴ y = u + v
Differentiating w.r.t. x
$$\Rarr\space \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}\\\text{...(i)}\\\text{Now,\space u =}\bigg(x + \frac{1}{x}\bigg)^{x}$$
Taking log on both sides
$$\Rarr\space\text{log u =}\\\text{log}\bigg(x + \frac{1}{x}\bigg)^{x}\\\Rarr\space\text{log u =}\text{x log}\bigg(x + \frac{1}{x}\bigg)$$
[∵ log mn = n log m]
Differentiating w.r.t. x
$$\Rarr\space\frac{1}{u}\frac{du}{dx} = x\frac{d}{dx}\text{log}\bigg(x +\frac{1}{x}\bigg) + \\\text{log}\bigg(x +\frac{1}{x}\bigg)\frac{d}{dx}(x)$$
[Using chain rule]
$$\Rarr\space\frac{1}{u}\frac{du}{dx} =\frac{x×1}{\bigg(x +\frac{1}{x}\bigg)}\bigg[1 -\frac{1}{x^{2}}\bigg] +\\\text{log}\bigg(x + \frac{1}{x}\bigg)\\\Rarr\space\frac{1}{u}\frac{du}{dx} =\frac{x}{\frac{(x^{2}+1)}{x}}\bigg[\frac{x^{2}-1}{x^{2}}\bigg]+\\\text{log}\bigg(x + \frac{1}{x}\bigg)\\\Rarr\space\frac{1}{u}\frac{du}{dx} =\begin{bmatrix}\frac{(x^{2}-1)}{(x^{2}+1)} + \text{log}\bigg(x+\frac{1}{x}\bigg)\end{bmatrix}\\\Rarr\frac{du}{dx} = u\begin{bmatrix}\frac{(x^{2}-1)}{(x^{2}+1)} + \text{log}\bigg(x +\frac{1}{x}\bigg)\end{bmatrix}$$
$$\Rarr\space\frac{du}{dx} =\bigg(x +\frac{1}{x}\bigg)^{x}\\\begin{bmatrix}\frac{(x^{2}-1)}{(x^{2}+1)} + \text{log}\bigg(x + \frac{1}{x}\bigg)\end{bmatrix}\\\text{Now, v = x}^{\bigg(1 + \frac{1}{x}\bigg)}$$
Taking log on both sides,
$$\Rarr\space\text{log v = log x}^{\bigg(x + \frac{1}{x}\bigg)}\\\Rarr\space\text{log v =}\bigg(1 + \frac{1}{x}\bigg)\text{log x}$$
Differentiating w.r.t. x,
$$\Rarr\space\frac{1}{v}\frac{dv}{dx} =\bigg(1 +\frac{1}{x}\bigg)\frac{d}{dx}(\text{log x}) + \\\text{log x}\frac{d}{dx}\bigg(1 + \frac{1}{x}\bigg)$$
(Using chain rule)
$$\Rarr\space\frac{1}{v}\frac{dv}{dx} =\bigg(1 + \frac{1}{x}\bigg)×\frac{1}{x}+\\\text{log x}\bigg[0 - \frac{1}{x^{2}}\bigg]\\\Rarr\space\frac{1}{v}\frac{dv}{dx} =\bigg[\frac{x+1}{x^{2}}\bigg] -\frac{1}{x^{2}}\text{log x}\\\Rarr\space\frac{1}{v}\frac{dv}{dx} =\bigg[\frac{x+1 -log x}{x^{2}}\bigg]\\\Rarr\space\frac{dv}{dx} = v\bigg[\frac{x+1 -log x}{x^{2}}\bigg]\\\Rarr\space\frac{dv}{dx}= x^{\bigg(x + \frac{1}{x}\bigg)}\bigg[\frac{x+1 - log x}{x^{2}}\bigg]$$
Putting the values of
$$\frac{du}{dx}\text{and}\frac{dv}{dx}\space\text{in Eq. (i),}$$
$$\therefore\space\frac{dy}{dx} = \bigg(x +\frac{1}{x}\bigg)^{x}\\\bigg[\frac{(x^{2}-1)}{(x^{2}+1)} + \text{log}\bigg(x + \frac{1}{x}\bigg)\bigg]+\\x^{\bigg(1 +\frac{1}{x}\bigg)}\bigg[\frac{x+1-log x}{x^{2}}\bigg]$$
7. (log x)x + xlog x.
Sol. Let y = (log x)x + xlog x
Let u = (log x)x
v = xlog x
∴ y = u + v
Differentiating w.r.t. x,
$$\frac{dy}{dx} =\frac{du}{dx} + \frac{dv}{dx}\\\text{...(i)}$$
Now, u = (log x)x
Taking log on both sides,
⇒ log u = log (log x)x
⇒ log u = x log (log x)
Differentiating w.r.t., x we get
$$\Rarr\space\frac{1}{u}\frac{du}{dx} = x\frac{d}{dx}\text{log(log x)}+ \\\text{log(log x)}\frac{d}{dx}(x) $$
(Using chain rule)
$$\Rarr\space\frac{1}{u}\frac{du}{dx} = \frac{x}{log x}× \frac{1}{x}+\\\text{log(log x)}\\\Rarr\space \frac{du}{dx} = u\bigg[\frac{1}{\text{log x}} + \text{log(log x)}\bigg]\\\Rarr\space\frac{du}{dx} =(\text{log x})^{x}\begin{bmatrix}\frac{1}{\text{log x}} +\\\text{log(log x)}\end{bmatrix}$$
Again, v = xlog x
Taking log on both sides,
⇒ log v = log xlog x
⇒ log v = (log x)(log x)
⇒ log v = (log x)2
Differentiating w.r.t. x, we get
$$\Rarr\space\frac{1}{v}\frac{dv}{dx} =\text{2 log x}\frac{d}{dx}(log x)\\= \text{2 log x}×\frac{1}{x}\\\bigg(\because\space\frac{d}{dx}[f(x)]^{2}=2f(x)\frac{d}{dx}f(x)\bigg)\\\Rarr\space\frac{dv}{dx} = v\bigg[\frac{2log x}{x}\bigg]\\\Rarr\space\frac{dv}{dx} = x^{log x}\bigg[\frac{2 log x}{x}\bigg]$$
Now, putting the values of
$$\frac{du}{dx}\text{and}\frac{dv}{dx}\space\text{in Eq.}$$
(i), we get
$$\frac{dy}{dx} =(\text{log x})^{x}\\\bigg[\frac{1}{\text{log x}} + \text{log(log x)}\bigg]+\\x^{\text{log x}}\bigg[\frac{\text{2 log x}}{x}\bigg]$$
$$\frac{dy}{dx} = (\text{log x})^{x-1}[1 +\text{log x}\\log(log x)] + 2x^{log x-1}log x$$
$$\textbf{8.\space (sin x)}^{\textbf{x}} \textbf{+}\textbf{sin}^{\normalsize-1}\sqrt{\textbf{x}}.\\\textbf{Sol.\space}y =(sin x)^{x} +\text{sin}^{\normalsize-1}\sqrt{x}\\\text{Let u = (sn x)}^{x},\\ v =\text{sin}^{\normalsize-1}\sqrt{x} $$
∴ v = u + v
Differentiating w.r.t., x
$$\Rarr\space\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}\\\text{...(i)}$$
Now, u = (sin x)x
Taking log on both sides,
log u = log (sin x)x ⇒ log u = x log (sin x)
Differentiating w.r.t., x we get
$$\Rarr\space\frac{1}{u}\frac{du}{dx} = x\frac{d}{dx}\text{log(sin x) +}\\ \text{log(sin x)}\frac{d}{dx}(x)$$
(Using product rule)
$$=\frac{x}{\text{sin x}}\space\text{(cos x + log(sin x))}\\\frac{du}{dx} =u\lbrack \text{x cot x + log sin x}\rbrack$$
= (sin x)x [x cot x + log sin x]
$$\text{Again,\space} v = \text{sin}^{\normalsize-1}\sqrt{x}$$
Differentiating w.r.t. x, we get
$$\text{Now,\space} v =\text{sin}^{\normalsize-1}\sqrt{x}\\\frac{dv}{dx} =\frac{1}{\sqrt{1 -(\sqrt{x})^{2}}}\frac{d}{dx}x^{1/2}\\=\frac{1}{\sqrt{1 - x}}\frac{1}{2}x^{\frac{1}{2}}$$
(Using chain rule)
$$\Rarr\space\frac{dv}{dx} =\frac{1}{\sqrt{1 - x}}×\frac{1}{2\sqrt{x}}\\\Rarr\space\frac{dv}{dx} =\frac{1}{2\sqrt{x}\sqrt{1- x}}\\\Rarr\space\frac{dv}{dx} =\frac{1}{2\sqrt{x - x^{2}}}$$
Putting the values of
$$\frac{du}{dx}\text{and}\frac{dv}{dx}\space\text{in Eq. (i), we get}\\\frac{dy}{dx} =(\text sin \space x)^{x}\\\lbrack\text{x cot x + log sinn x}\rbrack+\\\frac{1}{2\sqrt{x - x^{2}}}$$
9. xsin x + (sin x)cos x
Sol. Let y = xsin x + (sin xcos x)
Let u = xsin x, v= sin xcos x
∴ y = u + v
Differentiating w.r.t. x
$$\Rarr\space \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}\\\text{...(i)}$$
Now, u = xsin x
Taking log on both sides, log u = (sin x) log x
Differentiating w.r.t. x,
$$\frac{d}{dx}(\text{log u}) =\text{sin x}\frac{d}{dx}(\text{log x}) +\\\text{log x}\frac{d}{dx}(\text{sin x})$$
(∴ Using the product rule)
$$\Rarr\space\frac{1}{u}\frac{du}{dx} =\\\bigg[\text{sin x}×\frac{1}{x} + \text{log x cos x}\bigg]\\\Rarr\space\frac{du}{dx} = u\\\bigg[\frac{sin \space x}{x} + \text{cos x log x}\bigg]\\\Rarr\space\frac{du}{dx} = x^{sin x}\\\bigg[\frac{sin \space x}{x} +\text{cos x log x}\bigg]$$
Now, v = sin xcos x
Taking log on both sides, log v = cos x log(sin x)
Differentiating w.r.t. x,
$$\frac{d}{dx}(log \space v) =\text{cos x}\frac{d}{dx}\text{log (sin x) +}\\\text{log sin x}\frac{d}{dx}(\text{cos x})\\\Rarr\space\frac{1}{v}\frac{dv}{dx} =\begin{bmatrix}\text{cos x}×\frac{1}{\text{sin x}}×\text{cos x} +\\\text{log sinx(\normalsize-sin x)}\end{bmatrix}\\\Rarr\space\frac{dv}{dx} = v\lbrack\text{cot x cos x}-\\\text{sin xlog(sin x)}\rbrack\\\Rarr\space\frac{dv}{dx} =\text{sin x}^{\text{cos x}}\\\lbrack\text{cot x cos x - sin xlog(sin x)}\rbrack$$
Now, putting the values of
$$\frac{du}{dx}\text{and}\space \frac{dv}{dx}\space\text{in Eq. (i)}\\\frac{dy}{dx}= x^{sin x}\bigg[\frac{sin x}{x} +\text{cos x log x}\bigg]+\\\text{sin x}^{\text{cos x}}$$
[cot x cos x – sin x log (sin x)]
$$\textbf{10.\space x}^{\textbf{x cos x}} + \frac{\textbf{x}^{\textbf{2}}\textbf{+1}}{\textbf{x}^{\textbf{2}}\textbf{- 1}}\\\textbf{Sol.\space}\text{Let y = x}^{xcos x} +\\ \frac{x^{2}+1}{x^{2}-1}\\\text{Let }\space u = x^{x cos x},\\ v =\frac{x^{2}+1}{x^{2}-1}$$
∴ y = u + v
Differentiating w.r.t. x,
$$\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} $$
Now, u = xx cos x
Taking log on both sides for u, log u = x cos x log x
Differentiating w.r.t. x,
$$\frac{d}{dx}(\text{log u}) = xcos x\frac{d}{dx}(\text{log x}) +\\\text{log x}\frac{d}{dx}(x \space cos x)$$
$$= \text{x cos x}×\frac{1}{x} +\text{log x}\\\bigg[x\frac{d}{dx}\text{cos x + cos}x\frac{d}{dx}(x)\bigg]\\\Rarr\space\frac{1}{u}\frac{du}{dx} = \text{x cos x}×\frac{1}{x} +\\\text{log x}[(-\text{x sin x}) + \text{cos x}]\\\Rarr\space\frac{du}{dx} = u[\text{cos x - x sin xlog x +}\\ cos x.log x]\\\Rarr\space\frac{du}{dx} = x^{x cos x}\lbrack\text{cos x - xsin x log x}+\\\text{cos x.log x}\rbrack\\\text{Again,\space} v =\frac{x^{2}+1}{x^{2}-1}$$
Taking log on both sides,
log v = log (x2 + 1) – log(x2 – 1)
Differentiating w.r.t. x,
$$\Rarr\space\frac{d}{dx}\text{log v} =\frac{d}{dx}\text{log(x}^{2}+1)-\\\frac{d}{dx}\text{log}(x^{2}-1)\\\Rarr\space\frac{1}{v}\frac{dv}{dx} = \frac{1}{(x^{2}+1)}\frac{d}{dx}(x^{2}+1)-\\\frac{1}{(x^{2}-1)}\frac{d}{dx}(x^{2}-1)\\\Rarr\space\frac{1}{v}\frac{dv}{dx} = \frac{2x}{(x^{2}+1)} - \frac{2x}{(x^{2}-1)}$$
$$= 2x\begin{bmatrix}\frac{(x^{2}-1) - (x^{2}+1)}{(x^{2}+1)(x^{2}-1)}\end{bmatrix}\\\Rarr\space\frac{dv}{dx} = v\begin{bmatrix}\frac{\normalsize-4x}{x^{4}-1}\end{bmatrix}\\=\frac{x^{2}+1}{x^{2}-1}\begin{bmatrix}\frac{\normalsize-4x}{x^{4}-1}\end{bmatrix} =\frac{\normalsize-4x}{(x^{2}-1)^{2}}$$
Now, putting the values of
$$\frac{du}{dx}\space\text{and}\space\frac{dv}{dx}\space\text{in Eq. (i),}\\\therefore\space\frac{dy}{dx} = x^{x cos x}\\\lbrack cos \space x - \text{x sin x log x +}\\\text{cos x. log x}\rbrack +\\\frac{x^{2}+1}{x^{2}-1}\begin{bmatrix}\frac{\normalsize-4x}{x^{4}-1}\end{bmatrix} $$
$$= x^{x cos x}[cos x.(1 + log x)-\\\text{x sin x log x}-\frac{4x}{(x^{2}-1)^{2}}$$
$$\textbf{11.\space (x cos x)}^{\textbf{x}} \textbf{+}(\textbf{x sin x})^{\frac{\textbf{1}}{\textbf{x}}} .\\\textbf{Sol.\space}\text{Let y = }(x \space cos x)^{x} +\\ (x\space sin x)^{\frac{1}{x}},\\\text{Let u = (x cos x)}^{x},\\ v = (x\space sin x)^{\frac{1}{x}}$$
y = u + v
Differentiating w.r.t. x, we get
$$\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}\space\text{...(1)}$$
Now, u = (x cos x)x
Taking log on both sides, log u = x log (x cos x)
Differentiating w.r.t. x,
$$\frac{d}{dx}(\text{log u}) = x\frac{d}{dx}\text{log(x cos x)} +\\\text{log(x cos x)}\frac{d}{dx}(x)\\\Rarr\space\frac{1}{u}\frac{du}{dx} = x×\frac{1}{\text{x cosx}}\frac{d}{dx}\\(x \space cos x) +\text{log}(x cos x)×1\\\Rarr\space\frac{1}{u}\frac{du}{dx} = \frac{1}{\text{cos x}}\\\bigg(x \frac{d}{dx}\text{cos x + cos x}\frac{d}{dx}x\bigg)+\\\text{log(x cos x)}\\\Rarr\space\frac{1}{u}\frac{du}{dx} = \frac{1}{cos x}\\\lbrack x(-\text{sin x}) +\text{cos x}\rbrack +\text{log}(x\space cos x)$$
$$= \frac{1}{\text{cos x}}\lbrack\text{- x sin x + cos x}\rbrack +\\\text{log(x cos x)}\\\Rarr\space\frac{1}{u}\frac{du}{dx} = -\text{x tan x +1} + \\\text{log(x cos x)}\\\Rarr\space\frac{du}{dx} =u\lbrack-x \space tan x +1 +\\ \text{log(x cos x)}\rbrack\\\Rarr\space\frac{du}{dx} = (\text{x cos x})^{x}\lbrack-\text{x tan x +1 +}\\\text{log}(x\space cos x)\rbrack$$
$$\text{Now,\space v =} (x \space sin x)^{\frac{1}{x}}$$
Taking log on both sides,
$$\text{log v =}\frac{1}{x}\text{log(x sin x)}.$$
Differentiating w.r.t. x,
$$\frac{1}{v}\frac{d}{dx}\text{log v} =\frac{1}{x}\frac{d}{dx}(\text{log x sin x}) +\\\text{log(x sin x)}\frac{d}{dx}\bigg(\frac{1}{x}\bigg)\\\Rarr\space\frac{1}{v}\frac{dv}{dx} =\frac{1}{x}\\\begin{bmatrix}\frac{1}{\text{x sin x}}\frac{d}{dx}(\text{x sin x})\end{bmatrix} +\\\text{log(\text{x sin x})}\bigg(-\frac{1}{x^{2}}\bigg)$$
$$= \frac{1}{x}\begin{bmatrix}\frac{1}{\text{x sin x}}\begin{Bmatrix}x \frac{d}{dx}\text{sin x} +\\\text{sin x}\frac{d}{dx}x\end{Bmatrix}\end{bmatrix}-\\-\bigg(\frac{1}{x^{2}}\bigg)\text{log(x sin x)}$$
$$\Rarr\space\frac{1}{v}\frac{dv}{dx} =\frac{1}{x}×\frac{1}{\text{x sin x}}\\\lbrack\text{x cos x + sin x}\rbrack\\ + \text{log}(x \space sin x)\bigg(\frac{\normalsize-1}{x^{2}}\bigg)\\=\frac{\text{x cos x}}{x^{2}\text{sin x}} + \frac{\text{sin x}}{x^{2}\text{sin x}} +\\\bigg(\frac{\normalsize-1}{x^{2}}\bigg)\text{log x sin x}\\\Rarr\space \frac{dv}{dx} = v\begin{bmatrix}\frac{cot \space x}{x} + \frac{1}{x^{2}}-\\\frac{1}{x^{2}}\text{log x sin x}\end{bmatrix}\\\Rarr\space\frac{dv}{dx} = v\begin{bmatrix}\frac{\text{x cot x + 1 - log x sin x}}{x^{2}}\end{bmatrix}\\\Rarr\space\frac{dv}{dx} =(x \space sin x)^{\frac{1}{x}}$$
$$\begin{bmatrix}\frac{\text{x cot x + 1 - log x sin x}}{x^{2}}\end{bmatrix}$$
Now, putting the values of
$$\frac{du}{dx}\space\text{and}\space\frac{dv}{dx}\space\text{in Eq. (i),}\\\frac{dy}{dx} =(x \space cos x)^{x}\\\lbrack\text{-x tan x +1 + log}(x\space cos x)\rbrack\\ + (x\space sin x)^{\frac{1}{x}}\\\begin{bmatrix}\frac{\text{x cot x +1 - log x sin x}}{x^{2}}\end{bmatrix}$$
$$\textbf{Find}\space\frac{\textbf{dy}}{\textbf{dx}}\space\textbf{of the functions}\\\textbf{given in question.}$$
12. xy + yx = 1.
Sol. Given, xy + yx = 1
Let u = xy, v = yx
∴ u + v = 1
Differentiating w.r.t. x,
$$\frac{du}{dx} + \frac{dv}{dx} = 0\\\text{...(i)}$$
Now, u = xy
Taking log on both sides, we get log u = y log x
Differentiating w.r.t. x
$$\Rarr\space\frac{1}{u}\frac{du}{dx} = y×\frac{1}{x} +\\\text{log x}\frac{dy}{dx}\\\text{(Using product rule)}\\\Rarr\space\frac{du}{dx} = u\begin{bmatrix}\frac{y}{x} +\text{log x}\frac{dy}{dx}\end{bmatrix}\\\Rarr\space\frac{du}{dx} = x^{y}\begin{bmatrix}\frac{y}{x} +\text{log x}\frac{dy}{dx}\end{bmatrix}\\\Rarr\space\frac{du}{dx} = yx^{y-1} + \\x^{y}.log x\frac{dy}{dx}$$
Now, v = yx
Taking log on both sides
$$\Rarr\space\text{log v = x log y}$$
Differentiating w.r.t. x,
$$\Rarr\space\frac{1}{v}\frac{dv}{dx} = x×\frac{1}{y}\frac{dy}{dx} +\\\text{log y}\space (\text{Using product rule})\\\Rarr\space\frac{dv}{dx} = v\begin{bmatrix}\frac{x}{y}\frac{dy}{dx} + log\space y\end{bmatrix}\\\Rarr\space\frac{dv}{dx} = y^{x}\begin{bmatrix}\frac{x}{y}\frac{dy}{dx} + \text{log y}\end{bmatrix}\\\Rarr\space \frac{dv}{dx} = xy^{x-1}\frac{dy}{dx} + y^{x}\text{log y}$$
Now, putting the values of
$$\frac{du}{dx}\space\text{and}\space\frac{dv}{dx}\space\text{in Eq.(i)}.\\\text{yx}^{y-1} + x^{y}.\text{log x}\frac{dy}{dx} +\\ xy^{x-1}\frac{dy}{dx} + y^{x}\text{log y = 0}\\\Rarr\space \frac{dy}{dx}\lbrack x^{y}\text{log x + xy}^{x-1}\rbrack$$
= – yx log y – yxy – 1
$$\Rarr\space\frac{dy}{dx} =-\begin{bmatrix}\frac{y^{x}log\space y + yx^{y-1}}{x^{y}log x + xy^{x-1}}\end{bmatrix}$$
13. yx = xy.
Sol. Given, yx = xy
Taking log on both sides, we get log yx = log xy
⇒ x log y = y log x
Differentiating both sides w.r.t. x, we obtain
$$\frac{d}{dx}(\text{x} \space \text{log y}) = \frac{d}{dx}(\text{y log x})\\\Rarr\space x\bigg(\frac{1}{y}\bigg)\frac{dy}{dx} + \text{(log y)}\\ = y\frac{1}{x} +(\text{log x})\frac{dy}{dx}$$
(Using product rule)
$$\Rarr\space\frac{x}{y}\frac{dy}{dx} -\text{(log x)}\frac{dy}{dx}\\=\frac{y}{x} -\text{log y}\\\Rarr\space\bigg(\frac{x}{y} -\text{log x}\bigg)\frac{dy}{dx}\\=\frac{y}{x} -\text{log y}\\\Rarr\space\bigg(\frac{x -y\space log x}{y}\bigg)\frac{dy}{dx} =\\\frac{\text{y - x log y}}{x}\\\Rarr\space\frac{dy}{dx} = \frac{y}{x}\begin{pmatrix}\frac{\text{y - xlog y}}{\text{x - y log x}}\end{pmatrix}$$
14. (cos x)y + (cos y)x.
Sol. Given, (cos x)y = (cos y)x, taking log on both sides, we get
log {(cos x)y} = log {(cos y)x}
or y log (cos x) = x log (cos y)
Differentiating both sides w.r.t. x, we get
$$y\frac{d}{dx}(\text{log cos x}) + \text{log cos x}\frac{d}{dx}y\\ = x\frac{d}{dx}(\text{log cos y}) +\\ \text{log cos y}\frac{d}{dx} x$$
(Using product rule)
$$\Rarr\space y\bigg(\frac{1}{\text{cos x}}\bigg)(-\text{sin x}) + \\ \text{log (cos x)}\frac{dy}{dx}\\ = x\bigg(\frac{1}{\text{cos y}}\bigg)(-\text{sin y})\frac{dy}{dx} +\\\text{log}(cos y).1\\\Rarr\space\text{log(cos x)}\frac{dy}{dx} + \text{x tan y}\frac{dy}{dx}\\$$
= log (cos y) + y tan x
$$\Rarr\space (\text{log}(\text{cos} \space \text{x}) + \text{x tan y})\frac{dy}{dx}$$
= log (cos y) + y tan x
$$\Rarr\space\frac{dy}{dx} =\\\frac{\text{log}(cos \space y) + y(\text{tan x})}{\text{log}(cos x) + x \space tan y}$$
15. xy = e(x – y).
Sol. Given, xy = e(x – y)
Differentiating both sides w.r.t. x, we get
$$\frac{d}{dx}(xy) = \frac{d}{dx}(e^{x-y})\\\Rarr\space x\frac{dy}{dx} + y.1\\= e^{x-y}\frac{d}{dx}(x-y)$$
(Using product rule in LHS and chain rule in RHS)
$$\Rarr\space x\frac{dy}{dx} + y = e^{x-y}\bigg(1 -\frac{dy}{dx}\bigg)\\\Rarr\space x\frac{dy}{dx} + e^{x-y}\frac{dy}{dx} \\= e^{x-y}-y\\\Rarr\space (x + e^{x-y})\frac{dy}{dx} \\= e^{x-y}-y\\\Rarr\space\frac{dy}{dx} =\frac{e^{x-y}-y}{x + e^{x-y}}\\=\frac{xy-y}{x + xy}$$
(∵ ex – y = xy is given)
$$=\space \frac{y(x-1)}{x(1 +y)}$$
16. Find the derivative of the function given by f(x) = (1 + x)(1 + x2)(1 + x4)(1 + x8) and hence find f ′(1).
Sol. Given, f(x) = (1 + x)(1 + x2)(1 + x4)(1 + x8)
Taking log on both sides, we get
log (f(x)) = log {(1 + x)(1 + x2)(1 + x4)(1 + x8)}
⇒ log (f(x)) = log (1 + x) + log (1 + x2) + log(1 + x4) + log(1 + x8)
Differentiating both sides, w.r.t. x, we get
$$\frac{d}{dx}\text{log(f(x))} =\frac{d}{dx}\text{log(1 + x)} +\\\frac{d}{dx}\text{log(1 + x}^{2}) + \frac{d}{dx}\text{log(1+ x}^{4})+\\\frac{d}{dx}\text{log}(1 + x^{8})\\\frac{1}{f(x)}f'(x) =\frac{1}{\text{1-x}}\text{(1 - 0)} +\\\frac{1}{1 +x^{2}}(0 +2x) + \frac{1}{1 + x^{4}}(0 + 4x^{3}) +\\\frac{1}{1 +x^{8}}(0 + 8x^{7})\\\Rarr\space f'(x) = f(x)\\\begin{Bmatrix}\frac{1}{1 + x} + \frac{2}{1 + x^{2}} + \frac{4x^{3}}{1 +x^{4}}+ \frac{8x^{7}}{1+x^{8}}\end{Bmatrix} $$
= (1 + x)(1 + x2)(1 + x4)(1 + x8)
$$\begin{Bmatrix}\frac{1}{1 +x} + \frac{2x}{1 +x^{2}} + \frac{4x^{3}}{1 +x^{4}} + \frac{8x^{7}}{1 +x^{8}}\end{Bmatrix}$$
∴ f'(1) = (1 + 1)(1 + 1)(1 + 1)(1 + 1)
$$\begin{pmatrix}\frac{1}{1 + 1} + \frac{2}{1 + 1} + \frac{4}{1 + 1} +\frac{8}{1 + 1}\end{pmatrix}\\ = 2^{4}\begin{Bmatrix}\frac{1}{2} + 1 +2+4\end{Bmatrix}\\ = 16\bigg(\frac{15}{2}\bigg) = 120$$
17. Differentiate (x2 – 5x + 8)(x3 + 7x + 9) in three ways mentioned below.
(a) By using product rule.
(b) By expanding the product to obtain a single polynomial.
(c) By logarithmic differentiation.
Do they all given the same answer ?
Sol. Let y = (x2 – 5x + 8)(x3 + 7x + 9) ...(i)
(a) Using product rule,
$$\frac{dy}{dx} =(x^{2} - 5x+8)\frac{d}{dx}(x^{3} +7x +9)+\\(x^{3} +7x+ 9)\frac{d}{dx}(x^{2}- 5x + 8)$$
= (x2 – 5x + 8)(3x2 + 7) + (x3 + 7x + 9)(2x – 5)
= 3x4 – 15x3 + 24x2 + 7x2 – 35x + 56 + 2x4 + 14x2 + 18x – 5x3 – 35x – 45
= 5x4 – 20x3 + 45x2 – 52x + 11 ...(ii)
(b) Writing y as a single polynomial,
y = (x2 – 5x + 8)(x3 + 7x + 9)
= x5 – 5x4 + 8x3 + 7x3 – 35x2 + 56x + 9x2 – 45x + 72
= x5 – 5x4 + 15x3 – 26x2 + 11x + 72)
Differentiating both sides w.r.t. x, we get
$$\therefore\space\frac{dy}{dx} = \frac{d}{dx}(x^{5}-5x^{4} +15x^{3}\\-26x^{2} + 11x +72) $$
= 5x4 – 20x3 + 45x2 – 52x + 11 ...(iii)
(c) Taking logarithms on both sides of Eq. (i), we get
log y = log{(x2 – 5x + 8)(x3 + 7x + 9)}
⇒ log y = log (x2 – 5x + 8) + log(x3 + 7x + 9)
Differentiating both sides w.r.t. x, we get
$$\frac{1}{y}\frac{dy}{dx} =\frac{1}{x^{2}-5x +8}\\\frac{d}{dx}(x^{2}-5x+8) + \\\frac{1}{x^{3} +7x+9}\frac{d}{dx}(x^{3}+ 7x +9)$$
(Using chain rule)
$$\Rarr\space\frac{1}{y}\frac{dy}{dx} = \frac{2x-5}{x^{2}-5x+8} +\\\frac{3x^{2}+7}{x^{3} +7x+9}\\\Rarr\space\frac{dy}{dx} = y\begin{pmatrix}\frac{2x-5}{x^{2}- 5x +8}\end{pmatrix} +\\\frac{3x^{2} +7}{x^{3} +7x +9}\\\Rarr\space\frac{dy}{dx} = y\begin{pmatrix}\frac{2x-5}{x^{2}-5x+8} + \frac{3x^{2}+7}{x^{3} +7x +9}\end{pmatrix}$$
= (x2 – 5x + 8)(x3 + 7x + 9)
$$\begin{pmatrix}\frac{2x-5}{x^{2}-5x +8} + \frac{3x^{2} +7}{x^{3}+7x +9}\end{pmatrix}$$
= (2x – 5)(x3 + 7x + 9) + (3x2 + 7)(x2 – 5x + 8)
= 2x4 + 14x2 + 18x – 5x3 – 35x – 45
+ 3x4 – 15x3 + 24x2 + 7x2 – 35x + 56
= 5x4 – 20x3 + 45x2 – 52x + 11 ...(iv)
From equations (ii), (iii) and (iv), we find that result is same in every case.
18. If u, v and w are functions of x, then show that
$$\frac{\textbf{d}}{\textbf{dx}}(\textbf{u.v.w}) -\frac{\textbf{du}}{\textbf{dx}}\textbf{v.w +}\\\textbf{u.}\frac{\textbf{dv}}{\textbf{dx}}\textbf{.w +u.v}\frac{\textbf{dw}}{\textbf{dx}}$$
In two ways-first by repeated application of product rule, second by logarithmic differentiation.
Sol. First, we use product rule and find that
$$\frac{d}{dx}(uvw) =\frac{d}{dx}\lbrack(uv)w\rbrack =\\(uv)\frac{dw}{dx} + w\frac{d}{dx}(uv)$$
(Consider here uv as one function and w as another function)
$$= uv\frac{dw}{dx} + w\begin{pmatrix}u\frac{dv}{dx}+ v\frac{du}{dx}\end{pmatrix}\\= uv\frac{dw}{dx} + wu\frac{dv}{dx} + wv\frac{du}{dx}\\\text{...(i)} $$
Next, we use logarithmic differentiation to obtain
$$\frac{d}{dx}(uvw).$$
Let y = uvw ⇒ log y = log (uvw)
⇒ log y = log u + log v + log w
Differentiating w.r.t. x, we get
$$\frac{1}{y}\frac{dy}{dx} = \frac{1}{u}\frac{du}{dx} + \frac{1}{v}\frac{dv}{dx} +\frac{1}{w}\frac{dw}{dx}\\\Rarr\space\frac{dy}{dx} = y\\\begin{Bmatrix}\frac{1}{u}\frac{du}{dx} +\frac{1}{v}\frac{dv}{dx} + \frac{1}{w}\frac{dw}{dx}\end{Bmatrix}\\ =(uvw)\begin{Bmatrix}\frac{1}{u}\frac{du}{dx} +\frac{1}{v}\frac{dv}{dx} +\frac{1}{w}\frac{dw}{dx}\end{Bmatrix}\\=(vw)\frac{du}{dx} + (uw)\frac{dv}{dx} +(uv)\frac{dw}{dx}\\=uv\frac{dw}{dx} +uw\frac{dv}{dx} +wv\frac{du}{dx}\\\text{...(ii)}$$
From equations (i) and (ii), we find the result obained is same in the two cases.
