Vtyjkyuk

For this equation I'm using the following property
$$f(x)=e^kx$$
$$f'(x)=ke^kx$$

As well as the product rule
$$f(x)=uv$$
$$f'(x)=u'v+uv'$$

I factorize $x$ on $e$'s exponent and then use the first property to differentiate:
$$e^7x^3-frac53=e^x(7x^2)-frac53=7x^2*e^7x^3-frac53$$

Is his fully differentiated? Or do I have to apply the product rule to $7x^2$? Any other steps I'm missing?

We need to use chain rule

$$(e^f(x))'=f'(x)e^f(x)$$

with

$$f(x)=7x^3-frac53 implies f'(x)=21x^2$$

Remember that when differentiating exponential functions you always get a copy back, multiplied by some other stuff. So we know the derivative contains the term $expleft( 7x^3 - frac53 right)$.

Now we use the chain. The derivative of the power is $21x^2$ by the usual rules. Here, $frac53$ is a constant and so its derivative vanishes everywhere.

Thus, the derivative is $21x^2 cdot expleft( 7x^3 - frac53 right)$.

(Here $exp(x)$ means $e^x$.)

This is not quite right. It's true that
$$
fracmathrmdmathrmdxleft[ e^kx right] = ke^kx,,
$$
but this is because
$$
fracmathrmdmathrmdxleft[ e^f(x) right] = f'(x)cdot e^f(x),.
$$
Hence for your question you have
$$
fracmathrmdmathrmdxleft[ e^7x^3 - frac53 right] =
fracmathrmdmathrmdxleft[ 7x^3 - frac53 right]e^7x^3 - frac53,.
$$
Can you finish it from here?