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?
$\endgroup$ 33 Answers
$\begingroup$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 $e^x$.)
$\endgroup$ $\begingroup$We need to use chain rule
$$(e^{f(x)})'=f'(x)e^{f(x)}$$
with
$$f(x)=7x^3-\frac{5}{3} \implies f'(x)=21x^2$$
$\endgroup$ $\begingroup$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?
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