Question

How do you differentiate $y = {7^{{x^2}}}$?

Answer 1

In this problem we have given $y$ is equal to some integer with power and the power term is an unknown with some power. Here we are asked to differentiate the given term. For this first we need to take derivatives on both sides. Also we can differentiate this problem by using some important rules like exponent rule, the chain rule.

Formula used:
The exponent rule: ${a^b} = {e^{b\ln (a)}}$
The chain rule: $\dfrac{{df\left( u \right)}}{{dx}} = \dfrac{{df}}{{du}}\dfrac{{du}}{{dx}}$
Common derivative: $\dfrac{d}{{du}}\left( {{e^u}} \right) = {e^u}$
Taking the constant out: $\left( {af'} \right) = af'$
The power rule: $\dfrac{d}{{dx}}\left( {{x^a}} \right) = a.{x^{a - 1}}$
There are the formulas we are going to use for differentiating this problem.

Complete step by step answer:
Solving the derivative of $y = {7^{{x^2}}}$
Taking derivative on both sides, we get
$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {{7^{{x^2}}}} \right)$
Now let’s apply exponent rule and also$\dfrac{{dy}}{{dx}}$can be written as $y'$,
$y' = \dfrac{d}{{dx}}\left( {{e^{{x^2}\ln \left( 7 \right)}}} \right)$,
Here in the right hand side we cannot simply differentiate, so let us apply the chain rule method for the differentiation.
$\Rightarrow y' = \dfrac{d}{{dx}}\left( {{e^u}} \right)\dfrac{d}{{dx}}\left( {{x^2}\ln \left( 7 \right)} \right)$,
Next we want to take the constant term out from the differentiation. We get
$\Rightarrow y' = \dfrac{d}{{dx}}\left( {{e^u}} \right)\left( {\ln \left( 7 \right)\dfrac{d}{{dx}}\left( {{x^2}} \right)} \right) - - - - - (1)$,
$\ln \left( 7 \right)$ is the constant term, so we took it out.
Now, differentiating ${e^u}$ and ${x^2}$, we get
$\Rightarrow y' = {e^u}2\ln \left( 7 \right)x - - - - - (2)$
Now replace$u = {x^{2\ln \left( 7 \right)}}$in equation (2), we get
$\Rightarrow y' = {e^{{x^2}\ln \left( 7 \right)}}2\ln \left( 7 \right)x - - - - - (3)$
We can write${e^{{x^2}\ln \left( 7 \right)}}$as$7{x^2}$, then equation (3) becomes,
$\Rightarrow y' = {7^{{x^2}}}2\ln \left( 7 \right)x$, this is the derivation of a given term.

$\Rightarrow y' = {7^{{x^2}}}2\ln \left( 7 \right)x$ is the required solution.

Note: The derivation of an exponential function is a constant times itself. So in equation (1) the derivation of ${e^u}$ is again ${e^u}$. And also in equation (3) we have written ${e^{{x^2}\ln \left( 7 \right)}}$ as $7{x^2}$. Here we used the fact that the natural logarithm is the inverse of the exponential function, so ${e^{\ln \left( x \right)}} = x$, by logarithm identity $1$.