Question

Solve the given question ↑↑
Chapter - Differentiation
Class - 11 th
Thanks ^_^ ​

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

1.

$\displaystyle \: \frac{d}{dx} ( {x}^{n} ) = n \: {x}^{n - 1}$

2.$\displaystyle \: \frac{d}{dx} ( logx)= \frac{1}{x}$

3.$\displaystyle \: \frac{d}{dx} (uv \: ) =u \frac{dv}{dx} + v\frac{du}{dx}$

TO EVALUATE

Differentiate : $y = {x}^{2} \sqrt{x} + {x}^{4} \: logx$

EVALUATION

Let $u \: = {x}^{2} \sqrt{x} \: \: and \: \: v = {x}^{4} \: logx$

$\implies \: \displaystyle \: \frac{dv}{dx} = 4 {x}^{3} \: logx + {x}^{3}$

So $y = u + v$

Differentiating both sides with respect to x

$\displaystyle \: \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$

Now

$\displaystyle \: u \: = {x}^{2} \sqrt{x} = {x}^{(2 + \frac{1}{2} \: ) } = {x}^{ \frac{5}{2} }$

Differentiating both sides with respect to x we get

$\displaystyle \: \frac{du}{dx} = \frac{5}{2} {x}^{( \frac{5}{2} - 1)} = \frac{5}{2} {x}^{ \frac{3}{2} }$

Again $v = {x}^{4} \: logx$

Differentiating both sides with respect to x we get

$\displaystyle \: \frac{dv}{dx} = 4 {x}^{3} \: logx + {x}^{4} \times \frac{1}{x}$

RESULT

$\displaystyle \: \frac{dy}{dx} = \frac{5}{2} {x}^{ \frac{3}{2} } + 4 {x}^{3} \: logx \: + {x}^{3}$

Answer 2

I hope you know Implicit Differentiation.[Math Processing Error]

y=(x+2)(x−1)(x+3)(1)

Taking logarithms,

lny=ln(x+2)+ln(x−1)+ln(x+3)

Differentiating w.r.t x,

1y\dy\dx=1x+2+1x−1+1x+3

\dy\dx=y(1x+2+1x−1+1x+3)

From (1):

\dy\dx=(x−1)(x+3)+(x+2)(x+3)+(x+2)(x−1)

\dy\dx=(x−1)(x+3)+(x+2)(x+3+x−1)

\dy\dx=(x2+2x−3)+(2x2+6x+4)

\dy\dx=3x2+8x+1