$y=e^{ax^3}$
What in a problem indicates I should use the laws of logarithm?
Here is how I differentiated the function:
$\ln(y)=\ln{e^{ax^3}}$
$\ln(y)=ax^3\ln(e)$
$\frac{dy}{dx}\cdot\frac{1}{y}=3ax^2\cdot (1)$
$\frac{dy}{dx}\cdot\frac{1}{y}\cdot (y)=\ (y)\ 3ax^2\cdot (1)$
$\frac{dy}{dx}=\ e^{ax^3}\ 3ax^2$
$\endgroup$ 44 Answers
$\begingroup$You can do it like this, but I would just use the chain rule:
$(e^{ax^3})' = e^{ax^3} (ax^3)' = e^{ax^3} 3ax^2$.
More generally $(e^{f(x)})' = e^{f(x)} f'(x)$ for any differentiable function $f(x)$.
To differentiate $e^{-2t} \cos (4t)$, one can first use the product rule:
$$(e^{-2t} \cos (4t)) = (e^{-2t})' \cos (4t) + e^{-2t} \ (\cos (4t))'$$
then use the chain rule for $(e^{-2t})'$ and $(\cos (4t))'$.
$$y=e^{ax^3}$$ $$\ln y= \ln(e^{ax^3})$$ $$\ln y= ax^3$$ $$\frac 1y\frac{dy}{dx}=3ax^2$$ $$\frac{dy}{dx}=3ax^2\cdot y$$ $$\frac{dy}{dx}=3ax^2 e^{ax^3}$$
So your work is correct. Of course, this would be more easily done with the chain rule.
$\endgroup$ $\begingroup$Your calculus is correct. You can find the same result also using the chain rule: $$ \dfrac{d}{dx}e^{ax^3}= e^{ax^3}\dfrac{d}{dx}(ax^3)=3ax^2e^{ax^3} $$
$\endgroup$ $\begingroup$Hint:computational
$y=e^{f(x)}$then $\frac{dy}{dx}=e^{f(x)}f^1(x)$
$y=e^{ax^3}$
$=>\frac{dy}{dx}=e^{ax^3}\frac{d}{dx}(ax^3)=e^{ax^3}(3ax^2)$
$\endgroup$
