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