Differentiate $e^{7x^3-\frac{5}{3}}$ 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:
$$e^{7x^3-\frac{5}{3}}=e^{x(7x^2)-\frac{5}{3}}=7x^2*e^{7x^3-\frac{5}{3}}$$
Is his fully differentiated? Or do I have to apply the product rule to $7x^2$? Any other steps I'm missing?
 A: 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 $\exp\left( 7x^3 - \frac{5}{3} \right)$.
Now we use the chain. The derivative of the power is $21x^2$ by the usual rules. Here, $\frac{5}{3}$ is a constant and so its derivative vanishes everywhere.
Thus, the derivative is $21x^2 \cdot \exp\left( 7x^3 - \frac{5}{3} \right)$.
(Here $\exp(x)$ means $e^x$.)
A: We need to use chain rule
$$(e^{f(x)})'=f'(x)e^{f(x)}$$
with
$$f(x)=7x^3-\frac{5}{3} \implies f'(x)=21x^2$$
A: This is not quite right. It's true that
$$
\frac{\mathrm{d}}{\mathrm{d}x}\left[ e^{kx} \right] = ke^{kx}\,,
$$
but this is because
$$
\frac{\mathrm{d}}{\mathrm{d}x}\left[ e^{f(x)} \right] = f'(x)\cdot e^{f(x)}\,.
$$
Hence for your question you have
$$
\frac{\mathrm{d}}{\mathrm{d}x}\left[ e^{7x^{3} - \frac{5}{3}} \right] = 
\frac{\mathrm{d}}{\mathrm{d}x}\left[ 7x^{3} - \frac{5}{3} \right]e^{7x^{3} - \frac{5}{3}}\,.
$$
Can you finish it from here?
