Class 12 NCERT Solutions – Mathematics Part I – Chapter 5 Continuity And Differentiability – Exercise 5.5 | Set 2

Question 11. Differentiate the function with respect to x.

(x cos x)^x+ (x sin x)^1/x

Solution:

Given: (x cos x)^x+ (x sin x)^1/x
Let us considered y = u + v
Where, u = (x cos x)^xand v = (x sin x)^1/x
So, dy/dx = du/dx + dv/dx ………(1)
So first we take u = (x cos x)^x
On taking log on both sides, we get
log u = log(x cos x)^x
log u = xlog(x cos x)
Now, on differentiating w.r.t x, we get
$\frac{1}{u}\frac{du}{dx}=x\frac{d}{dx}(\log x+\log(\cos x))+\log x+\log \cos x$
$\frac{1}{u}\frac{du}{dx}=x(\frac{1}{x}+\frac{1}{\cos x}\frac{d}{dx}\cos x)+\log x+\log\cos x$
$\frac{du}{dx}=u(x(\frac{1}{x}+\frac{-\sin x}{\cos x})+\log x+\log(\cos x))$
$\frac{du}{dx}=(x\cos x)^x(1-x\tan x+\log x+\log(\cos x))$ ………(2)
Now we take u =(x sin x)^1/x
On taking log on both sides, we get
log v = log (x sin x)^1/x
log v = 1/x log (x sin x)
log v = 1/x(log x + log sin x)
Now, on differentiating w.r.t x, we get
$\frac{1}{v}\frac{dv}{dx}=\frac{1}{x}\frac{d}{dx}(\log x+\log(\sin x)+\frac{d}{dx}(\frac{1}{x}).(\log x+\log(\sin x)))$
$\frac{1}{v}.\frac{dv}{dx}=\frac{1}{x}(\frac{1}{x}+\frac{1}{\sin x}.\frac{d}{dx}\sin x)+(\frac{-1}{x^2})(\log x+\log(\sin x))$
$\frac{dv}{dx}=v(\frac{1}{x}(\frac{1}{x}+\frac{\cos x}{\sin x})\frac{-1}{x^2}(\log x+\log(\sin x)))$
$\frac{dv}{dx}=(x\sin x)^{1/2}.[(\frac{1}{x^2}+\frac{\cot x}{x})-\frac{\log x}{x^2}-\frac{\log(\sin x)}{x^2}]$ ………(3)
Now put all the values from eq(2) and (3) into eq(1)
$\frac{dy}{dx}=(x\cos)^x(1-x\tan x+\log x+\log(\cos x))+(x\sin x)^{\frac{1}{x}}.[\frac{xcotx+1-log(xsinx)}{x^2}]$

Find dy/dx of the function given in questions 12 to 15

Question 12. x^y+ y^x= 1

Solution:

Given: x^y+ y^x= 1
Let us considered
u = x^y and v = y^x
So,
$\frac{du}{dx}+\frac{dv}{dx}=0$ ………(1)
So first we take u = x^y
On taking log on both sides, we get
log u = log(x^y)
log u = y log x
Now, on differentiating w.r.t x, we get
$\frac{1}{u}.\frac{du}{dx}=y.\frac{d}{dx}\log x+\frac{dy}{dx}.\log x$
$\frac{1}{u}\frac{du}{dx}=\frac{y}{x}+\frac{dy}{dx}\log x$
$\frac{du}{dx}=x^4(\frac{y}{x}+\frac{dy}{dx}\log x)$ ………(2)
Now we take v = y^x
On taking log on both sides, we get
log v = log(y)^x
log v = x log y
Now, on differentiating w.r.t x, we get
$\frac{1}{v}.\frac{dv}{dx}=x\frac{d}{dx}(\log x)+\log y\frac{d}{dx}x$
$\frac{dv}{dx}=v(x.\frac{1}{y}.\frac{dy}{dx}+\log y)$
$\frac{dv}{dx}=y^x(\frac{x}{y}\frac{dy}{dx}+\log x)$ ………(3)
Now put all the values from eq(2) and (3) into eq(1)
$x^y(\frac{y}{x}+\frac{dy}{dx}\log x)+y^x(\frac{x}{y}\frac{dy}{dx}+\log y)=0$
$(x^y.\log x+xy^{x-1})\frac{dy}{dx}=-(yx^{y-1}+y^x\log y)$
$\frac{dy}{dx}=\frac{-yx^{y-1}+y^x\log y}{x^y\log x+xy^{x-1}}$

Question 13. y^x= x^y

Solution:

Given: y^x= x^y
On taking log on both sides, we get
log(yx) = log(x^y)
xlog y = y log x
Now, on differentiating w.r.t x, we get
$x\frac{dy}{dx}(\log y)+\log y(\frac{d}{dx}x)=y\frac{d}{dx}\log x+\log x\frac{d}{dx}y$
$x.\frac{d}{dx}.y+\log y.1=y.\frac{1}{x}+\log x\frac{dy}{dx}$
$\frac{x}{y}\frac{dy}{dx}+\log y=\frac{y}{x}+\log x\frac{dy}{dx}$
$(\frac{x}{y}-\log x)\frac{dy}{dx}=(\frac{y}{x}-\log y)$
$\frac{dy}{dx}=\frac{\frac{y}{x}-\log y}{\frac{x}{y}-\log x}$
$\frac{dy}{dx}=\frac{y}{x}(\frac{y-x\log y}{x-y\log x})$

Question 14. (cos x)^y= (cos y)^x

Solution:

Given: (cos x)^y= (cos y)^x
On taking log on both sides, we get
y log(cos x) = x log (cos y)
Now, on differentiating w.r.t x, we get
$y\frac{d}{dx}\log(\cos x)+\log(\cos x).\frac{dy}{dx}=x\frac{d}{dx}\log (\cos y)+\log(\cos y)\frac{dx}{dx}$
$y\frac{1}{\cos x}\frac{d}{dx}\cos x+\log(\cos x)\frac{dy}{dx}=x\frac{1}{\cos y}\frac{d}{dx}\cos y+\log(\cos y).1$
$\frac{y}{\cos x}.-\sin x+\log(\cos x).\frac{dy}{dx}=\frac{x}{\cos y}.(-\sin y).\frac{dy}{dx}+\log(\cos y)$
$(\log(\cos x)+x\tan y)\frac{dy}{dx}=\log(\cos y)+y \tan x$
$\frac{dy}{dx}=\frac{\log(\cos y)+y\tan x}{\log(\cos x)+x\tan y}$

Question 15. xy = e^{(x – y)}

Solution:

Given: xy = e^{(x – y)}
On taking log on both sides, we get
log(xy) = log e^{x – y}
log x + log y = x – y
Now, on differentiating w.r.t x, we get
$\frac{d}{dx}\log x+\frac{d}{dx}\log y=\frac{dx}{dx}-\frac{dy}{dx}$
$\frac{1}{x}.+\frac{1}{y}.\frac{dy}{dx}=1-\frac{dy}{dx}$
$(\frac{1}{y}+1)\frac{dy}{dx}=(1-\frac{1}{x})$
$\frac{dy}{dx}=\frac{(1-\frac{1}{x})}{(1+\frac{1}{y})}$
$\frac{dy}{dx}=\frac{y(x-1)}{x(y+1)}$

Question 16. Find the derivative of the function given by f(x) = (x + 1)(x + x²)(1 + x⁴)(1 + x⁸) and hence find f'(1).

Solution:

Given: f(x) = (x + 1)(x + x²)(1 + x⁴)(1 + x⁸)
Find: f'(1)
On taking log on both sides, we get
log(f(x)) = log(1 + x) + log(1 + x²) + log(1 + x⁴) + log(1 + x⁸)
Now, on differentiating w.r.t x, we get
$\frac{1}{f(x)}.\frac{d}{dx}f{x}=\frac{1}{1+x}\frac{d}{dx}(1+x)+\frac{1}{1+x^2}(1+x)^2+\frac{1}{1+x^4}.\frac{d}{dx}(1+x^4)+\frac{1}{1+x^8}\frac{d}{dx}(1+x^8)\frac{f'(x)}{f(x)}=\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\frac{8x^7}{1+x^8}$
$f'(x)=(1+x)(1+x^2)(1+x^4)(1+x^8)(\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4x^3}{1+x^2}+\frac{8x^7}{1+x^8})$
∴ f'(1) = 2.2.2.2. $(\frac{1}{2}+\frac{2}{2}+\frac{4}{2}+\frac{8}{2})$
$f'(1)=16.(\frac{15}{2})$
f'(1) = 120

Question 17. Differentiate (x⁵– 5x + 8)(x³+ 7x + 9) in three ways mentioned below

(i) By using product rule

(ii) By expanding the product to obtain a single polynomial

(iii) By logarithmic differentiation.

Do they all give the same answer?

Solution:

(i) By using product rule
$\frac{d}{dx}(u.v)=v\frac{du}{dv}+u\frac{dv}{dx}$
$\frac{dy}{dx}=(x^2-5x+8)\frac{d}{dx}(x^3+7x+9)+(x^3+7x+9).\frac{d}{dx}(x^2-5x+8)$
dy/dx = (3x⁴– 15x³+ 24x²+ 7x²– 35x + 56) + (2x⁴+ 14x²+ 18x – 5x³– 35x – 45)
dy/dx = 5x⁴– 20x³+ 45x²– 52x + 11
(ii) By expansion
y = (x²– 5x + 8)(x³+ 7x + 9)
y = x⁵+ 7x³+ 9x²– 5x⁴– 35x²– 45x + 8x³+ 56x + 72
y = x⁵– 5x⁴+ 15x³– 26x²+ 11x + 72
dy/dx = 5x⁴– 20x³+ 45x²– 52x + 11
(iii) By logarithmic expansion
Taking log on both sides
log y = log(x²– 5x + 8) + log(x³+ 7x + 9)
Now on differentiating w.r.t. x, we get
$\frac{1}{y}.\frac{dy}{dx}=\frac{1}{x^2-5x+8}.\frac{d}{dx}(x^2-5x+8)+\frac{1}{x^3+7x+9}\frac{d}{dx}(x^3+7x+9)$
$\frac{1}{y}.\frac{dy}{dx}=\frac{2x-5}{x^2-5x+8}+\frac{3x^2+7}{x^3+7x+9}$
$\frac{1}{(x^2-5x+8)(x^3+7x+9)}\frac{dy}{dx}=\frac{(2x-5)(x^3+7x+9)+(3x^2+7)(x^2-5x+8)}{(x^2-5x+8)(x^3+7x+9)}$
dy/dx = 2x⁴+ 14x²+ 18x – 5x³– 35x – 45 + 3x⁴– 15x³+ 24x²+ 7x²– 35x + 56
dy/dx = 5x⁴– 20x³+ 45x²– 52x + 11
Answer is always same what-so-ever method we use.

Question 18. If u, v and w are function of x, then show that

$\frac{d}{dx}(u.v.w)=\frac{du}{dx}v.w+u.\frac{dv}{dx}.w+u.v.\frac{dw}{dx}$

Solution:

Let y = u.v.w.
Method 1: Using product Rule
$\frac{dy}{dx}=u\frac{d}{dx}(v.w)+v.w\frac{d}{dx}u$
$\frac{dy}{dx}=u.[v.\frac{dw}{dx}+w\frac{du}{dx}]+v.w.\frac{du}{dx}$
$\frac{dy}{dx}=u.v.\frac{dw}{dx}+u.w.\frac{dv}{dx}+v.w\frac{du}{dx}$
Method 2: Using logarithmic differentiation
Taking log on both sides
log y = log u + log v + log w
Now, Differentiating w.r.t. x
$\frac{1}{y}\frac{dy}{dx}=\frac{1}{u}\frac{du}{dx}+\frac{1}{v}\frac{dv}{dx}+\frac{1}{w}.\frac{dw}{dx}$
$\frac{dy}{dx}=(u.v.w)(\frac{1}{u}\frac{du}{dx}+\frac{1}{v}\frac{dv}{dx}+\frac{1}{w}\frac{dw}{dx})$
$\frac{dy}{dx}=v.w\frac{du}{dx}+uw\frac{dv}{dx}+uv\frac{dw}{dx}$