Htykuut

What am I doing wrong?

I am using the chain rule:
$$dfrac{d[f(g(x))]}{dx} = f'(g(x)) times g'(x)$$

So in our case:

$$f(x)=e^{x} quadtext{(outer function)}$$
$$g(x)=2x quadtext{(inner function)}$$
$$f'(x)=e^x $$
$$g'(x)=2$$

now:
$$dfrac{d[e^2x]}{dx} = f'(g(x)) times g'(x) = 2e^2$$

The answer is $2e^{2x}$

What went wrong?

Since $f'(x)=e^x$, $g(x)=2x$, and $g'(x)=2$, $f'bigl(g(x)bigr)g'(x)=e^{2x}times2=2e^{2x}$.

$g(x) = 2x; g'(x) = 2$ and $f(x) = f'(x) = e^x$

$dfrac{d[e^{2x}]}{dx} = f'(g(x)) times g'(x)=$

$f'(2x)times 2=$

$e^{2x}times 2 = 2e^{2x}$.

I can only surmise that you calculated

$f'(g'(x))g'(x) = 2e^2$ instead of $f'(g(x))g'(x) = 2e^{2x}$.

Put $f(x) = e^x$, $g(x) = 2x$. Then $f(g(x)) = e^{2x}$. Apply the chain rule as follows
$$(fcirc g)'(x) = f'(g(x))g'(x) = e^{2x}cdot 2 = 2e^{2x}.$$

For $g(x)$, you plugged in $g'(x)$. This means the exponent becomes 2 instead of $2x$, which is what's wrong.

You can think of $e^{2x}$ as $e^u$, where $u=2x$.

The chain rule can be written without the primes to avoid confusion as to $f'(g(x))g'(x)$.

$frac{d(e^{2x})}{dx}=frac{de^u}{dx}=frac{de^u}{du}frac{du}{dx}$

$frac{de^u}{du}=e^u=e^{2x}$ since $u=2x$.

$u=2x$, so $frac{du}{dx}=2$

So :

$frac{d(e^{2x})}{dx}=frac{de^u}{dx}=frac{de^u}{du}frac{du}{dx}=2e^{2x}$