Question

differentiate: e^ax sin^2x cos^3x
dy/dx=?

given ans is : e^ax cos^3x sin^2x (a-3tanx+2cotx)
pls help not getting right....

Answer 1

try logrithmic differentiation for this.

Answer 2

Yah Logrithmic Differentiation might be a good way to go.
You could also apply the Product Rule. Here is how that would look:
\[\large y=e^{ax}\cdot \sin^2x\cdot \cos^3x\]
\[\large \begin{align*}y'&=\color{royalblue}{(e^{ax})'}\cdot \sin^2x\cdot \cos^3x\\&+e^{ax}\cdot \color{royalblue}{(\sin^2x)'}\cdot \cos^3x\\&+e^{ax}\cdot \sin^2x\cdot \color{royalblue}{(\cos^3x)'}\end{align*}\]
\[\large \begin{align*}y'&=\color{orangered}{(ae^{ax})}\cdot \sin^2x\cdot \cos^3x\\&+e^{ax}\cdot \color{orangered}{(2\sin x \cos x)}\cdot \cos^3x\\&+e^{ax}\cdot \sin^2x\cdot \color{orangered}{(-3\cos^2x \sin x)}\end{align*}\]

If we factor a sin^2x out of each term, it will give us a sin x in the denominator of the second term because we don't quite have enough of them.
If we factor a cos^3x out of each term, it will give us a cos x in the denominator of the third term for the same reason, we only have a cos^2x there.

\[\large y'=e^{ax}\cdot \sin^2x\cdot \cos^3x \left[a+2\frac{\cos x}{\sin x}-3\frac{\sin x}{\cos x}\right]\]