Differentiate the function x^ x^ 2 -3 +(x-3)^ x^ 2 , for x>3 w.r.t to x.

Let y = x^ x^ 2 -3 +(x-3)^ x^ 2 And let u = x^ x^ 2 -3 and v = (x-3)^ x^ 2 y = u + v Differentiating both sides w.r.t. x we get d y d x =d u d x +d v d x .....(i) Now, u = x^ x^ 2 -3 Taking logarithm both sides log u = log x^ x^ 2 -3 log u = (x2 - 3) log x Differentiating w.r.t. x, we get 1 u du dx = _ X d dx (x^ 2 -3 )+ (x^ 2 -3 ) d dx ( x) d u d x =u [ x 2 x+ (x^ 2 -3 ) 1 x ] d u d x =x^ x^ 2 -3 [ x^ 2 -3 x +2 x x ].....(ii) Also, v = (x-3)^ x^ 2 Taking logarithm both sides log v = log (x-3)^ x^ 2 log v = x2log(x-3) Differentiating both sides w.r.t. x 1 v dv dx = (x-3) d dx (x^ 2 )+x^ 2 d dx [ (x-3)] d v d x =v [ (x-3) 2 x+x^ 2 1 (x-3) d d x (x-3) ] d v d x =(x-3)^ x^ 2 [2 x (x-3)+ x^ 2 (x-3) 1 ] d v d x =(x-3)^ x^ 2 [ x^ 2 (x-3) +2 x (x-3) ]....(iii) Substituting (ii) and (iii) in (i) d y d x =x^ x^ 2 -3 [ x^ 2 -3 x +2 x x ] + (x-3)^ x^ 2 [ x^ 2 (x-3) +2 x (x-3) ]

Differentiate the function x^ x^ 2 -3 +(x-3)^ x^ 2 , for x>3 w.r.…

Download App

Question

Differentiate the function $x^{x^{2}-3}+(x-3)^{x^{2}}, \text { for } x>3$ w.r.t to x.

✓

Answer

Let y = $x^{x^{2}-3}+(x-3)^{x^{2}}$
And let u = $x^{x^{2}-3}$ and v = $(x-3)^{x^{2}}$
$\therefore $ y = u + v
Differentiating both sides w.r.t. x we get
$\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}$.....(i)
Now,
u = $x^{x^{2}-3}$
Taking logarithm both sides
log u = log $x^{x^{2}-3}$
$\Rightarrow$ log u = (x² - 3) log x
Differentiating w.r.t. x, we get
$\frac{1}{\mathrm{u}} \frac{\mathrm{du}}{\mathrm{dx}}=\log _{\mathrm{X}} \times \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{2}-3\right)+\left(\mathrm{x}^{2}-3\right) \times \frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{x})$
$\Rightarrow \frac{d u}{d x}=u\left[\log x \times 2 x+\left(x^{2}-3\right) \times \frac{1}{x}\right]$
$\Rightarrow \frac{d u}{d x}=x^{x^{2}-3}\left[\frac{x^{2}-3}{x}+2 x \log x\right]$.....(ii)
Also,
v = $(x-3)^{x^{2}}$
Taking logarithm both sides
log v = log $(x-3)^{x^{2}}$
$\Rightarrow$log v = x²log(x-3)
Differentiating both sides w.r.t. x
$\frac{1}{\mathrm{v}} \frac{\mathrm{dv}}{\mathrm{dx}}=\log (\mathrm{x}-3) \times \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{2}\right)+\mathrm{x}^{2} \times \frac{\mathrm{d}}{\mathrm{dx}}[\log (\mathrm{x}-3)]$
$\Rightarrow \frac{d v}{d x}=v\left[\log (x-3) \times 2 x+x^{2} \times \frac{1}{(x-3)} \times \frac{d}{d x}(x-3)\right]$
$\Rightarrow \frac{d v}{d x}=(x-3)^{x^{2}}\left[2 x \log (x-3)+\frac{x^{2}}{(x-3)} \times 1\right]$
$\Rightarrow \frac{d v}{d x}=(x-3)^{x^{2}}\left[\frac{x^{2}}{(x-3)}+2 x \log (x-3)\right]$....(iii)
Substituting (ii) and (iii) in (i)
$\therefore \frac{d y}{d x}=x^{x^{2}-3}\left[\frac{x^{2}-3}{x}+2 x \log x\right]$ + $(x-3)^{x^{2}}\left[\frac{x^{2}}{(x-3)}+2 x \log (x-3)\right]$