Question

question
find the derivative of the following function.
y = e^{-7x^{5}-7x^{4}}

Explanation:

Step1: Recall chain - rule and power - rule

The derivative of a sum/difference of functions is the sum/difference of their derivatives, i.e., if $y = u - v$, then $y'=u' - v'$. Also, the chain - rule states that if $y = e^{u}$, then $y'=e^{u}\cdot u'$, and the power - rule states that if $y = ax^{n}$, then $y'=nax^{n - 1}$. Let $u = e^{-7x^{5}}$ and $v = 7x^{4}$.

Step2: Find the derivative of $u = e^{-7x^{5}}$

Let $t=-7x^{5}$. Then $u = e^{t}$. By the chain - rule, $u'=\frac{du}{dt}\cdot\frac{dt}{dx}$. We know that $\frac{du}{dt}=e^{t}$ and $\frac{dt}{dx}=-35x^{4}$. Substituting $t = - 7x^{5}$ back in, we get $u'=e^{-7x^{5}}\cdot(-35x^{4})=-35x^{4}e^{-7x^{5}}$.

Step3: Find the derivative of $v = 7x^{4}$

Using the power - rule, if $v = 7x^{4}$, then $v'=4\times7x^{4 - 1}=28x^{3}$.

Step4: Find the derivative of $y$

Since $y = u - v$, then $y'=u' - v'$. Substituting the values of $u'$ and $v'$ we found above, we have $y'=-35x^{4}e^{-7x^{5}}-28x^{3}$.

Answer:

$y'=-35x^{4}e^{-7x^{5}}-28x^{3}$