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?
Best Answer
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 $e^x$.)