Question

Find derivative of $y = {x^x}.{e^{2x + 5}}$.

Answer 1

Hint: Differentiation- It is the action of computing a derivative.
The derivative of a function \[y = f\left( x \right)\] of a variable x is a measure of the rate at which the value y of the function changes with respect to the change of the variable x.
It is denoted by \[dy/dx.\]
Some formulae of finding differentiation
$\dfrac{d}{{dx}}(ax) = a$
$\dfrac{d}{{dx}}(x) = 1$
$\dfrac{d}{{dx}}(c) = 0$ \[\left[ {c = constant} \right]\]
$\dfrac{d}{{dx}}({x^n}) = {x^{n - 1}}$
$\dfrac{d}{{dx}}({e^n}) = {e^x}$
$\dfrac{d}{{dx}}(ax \pm b) = \dfrac{d}{{dx}}(ax) \pm \dfrac{d}{{dx}}(b)$
$\dfrac{d}{{dx}}(\log x) = \dfrac{1}{x}$
$\dfrac{d}{{dx}}(\sin x) = \cos x$
$\dfrac{d}{{dx}}(cosx) = - \sin x$
$\dfrac{d}{{dx}}(\tan x) = \sec {x^2}$
$\dfrac{d}{{dx}}(\sec x) = \sec x\tan x$
$\dfrac{d}{{dx}}(\cot x) = - \cos e{c^2}x$
$\dfrac{d}{{dx}}(\cos ecx) = \cos ecx\cot x$

Complete step by step solution:
$y = {x^x}{e^{2x + 5}}$
Taking log both the sides
$\log y = \log ({x^x}{e^{2x + 5}})$
or
$\log y = \log {x^x} + \log {e^{2x + 5}}$
Differentiating both the sides
$\dfrac{d}{{dx}}(\log y) = \dfrac{d}{{dx}}(\log {x^x}) + \dfrac{d}{{dx}}[\log {e^{2x + 5}}]$
$\dfrac{1}{y}\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}(x\log x) + \dfrac{d}{{dx}}[(2x + 5){\log _e}e]$
$\dfrac{1}{y}\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}(x) \times \log x + x\dfrac{d}{{dx}}(\log x) + \dfrac{d}{{dx}}(2x + 5)$
$\dfrac{1}{y}\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}(x) \times \log x + x\dfrac{d}{{dx}}(\log x) + \dfrac{d}{{dx}}(2x + 5)$
$\dfrac{1}{y}\dfrac{{dy}}{{dx}} = (1)\log x + x \times \left( {\dfrac{1}{x}} \right) + 2$
$\dfrac{1}{y}\dfrac{{dy}}{{dx}} = \log x + 1 + 2$
$\dfrac{{dy}}{{dx}} = (\log x + 3) \times y$
or $\dfrac{{dy}}{{dx}} = [\log x + 3][{x^x}{e^{2x + 5}}]$
So derivative of $y = {e^x}{e^{2x + 5}}$ is $[\log x + 3]({x^x}{e^{2x + 5}})$.

Note: 1.Students usually forget to use u.v formula i.e. differentiation by parts for finding derivation of two functions which are in multiplication with each other.
U.v formula \[\dfrac{{d(uv)}}{{dx}} = \dfrac{{v.d\left( u \right)}}{{dx}} + \dfrac{{u.d\left( v \right)}}{{dx}}\]
(differentiation by parts)
One function will remain constant and the 2nd function to be differentiated.
Then the 2nd function will remain constant and the 1st one is to be differentiated.