Question

nth derivative of e^3x. sin^2 x​

Answer 1

Answer:

let y=e^3x. sin^2 x

solution:r^n.e^x.cos[bx+c+n^2]

Answer 2

Answer:

The nth derivative of

$e^{3x} *sin^{2x}$ $=r^n.e^x.cos[bx+c+n^2]$.



Step-by-step explanation:

Given :

$e^{3x} *sin^{2x}$

To find the nth derivative of $e^{3x} *sin^{2x}$.

Step 1

Let $y=e^{3x} *sin^{2x}$

$y=e^{3 x}\left(\frac{1-\cos 2 x}{2}\right)$

$y=\frac{1}{2} e^{5 x}(1-\cos 2 x)$

$y_{}=\frac{1}{2}\left(e^{5 x}-e^{3 x} \cos 2 x\right)$

 $y_{n}=\frac{1}{2}\left(D^{n}\left[e^{3 x}\right]-D^{n}\left[e^{3 x} \cos x\right]\right)$$\Rightarrow y_{n}=\frac{e^{3 x}}{2} \cdot\left[3^{n}-13^{n / 2} \cos \left(2 x+n+a^{-1} 2 / 3\right)].\right.$

$D^{n}\left[e^{m x}\right]=m^{n} \cdot e^{m x}$

$D^{n}\left[e^{2 x} \cos b x+c\right]$

$=\gamma^{n} \cdot e^{x} \cdot \cos \left[b x+c+n^{2}\right]$

Step 2

where $r=\sqrt{a^{2}+b^{2}}$ and

$\theta \ =\tan ^{-1}\left[\frac{b}{a}\right]$

Therefore, the nth derivative of

$e^{3x} *sin^{2x}$ $=r^n.e^x.cos[bx+c+n^2]$.