How do you differentiate f(x) =((65e^-7x)+2)^3  using the chain rule?

Mar 3, 2018

$823875 {e}^{98} {x}^{2} + 50700 {e}^{49} x + 780 {e}^{-} 7$, or to make it simpler:

$195 {e}^{-} 7 {\left(\left(65 {e}^{-} 7 x\right) + 2\right)}^{2}$

Explanation:

We have to find:

$\frac{d}{\mathrm{dx}} {\left(\left(65 {e}^{-} 7 x\right) + 2\right)}^{3}$

We can say that this is in the form ${u}^{3}$, where $u = \left(65 {e}^{-} 7 x\right) + 2$.

To solve this, then, we must compute $\frac{d}{\mathrm{du}} {u}^{3}$ and multiply that by $\frac{d}{\mathrm{dx}} \left(\left(65 {e}^{-} 7 x\right) + 2\right)$

As $\frac{d}{\mathrm{dx}} {x}^{n} = n {x}^{n - 1}$, we have:

$3 {u}^{2} \cdot \frac{d}{\mathrm{dx}} \left(\left(65 {e}^{-} 7 x\right) + 2\right)$

$3 {u}^{2} \cdot \frac{d}{\mathrm{dx}} \left(65 {e}^{-} 7 x\right)$

$3 {u}^{2} \cdot 65 {e}^{-} 7 \cdot \frac{d}{\mathrm{dx}} x$

$3 {u}^{2} \cdot 65 {e}^{-} 7$

And as $u = \left(65 {e}^{-} 7 x\right) + 2$, we have:

$195 {e}^{-} 7 {\left(\left(65 {e}^{-} 7 x\right) + 2\right)}^{2}$

What I do after this is just expansion. It isn't needed, and it just makes it more confusing.

$195 {e}^{-} 7 \left({\left(65 {e}^{-} 7 x\right)}^{2} + 2 \left(2\right) \left(65 {e}^{-} 7 x\right) + {2}^{2}\right)$

$195 {e}^{-} 7 \left(4225 {e}^{-} 14 {x}^{2} + 260 {e}^{-} 7 x + 4\right)$

$823875 {e}^{98} {x}^{2} + 50700 {e}^{49} x + 780 {e}^{-} 7$