The derivative of e3x sin4x with respect to x, is

A
$5 e^{3 x} \sin (4 x + {Tan}^{- 1} \frac{4}{3})$
B
$5 e^{3 x} \sin (4 x - {Tan}^{- 1} \frac{4}{3})$
C
$5 e^{3 x} \sin (4 x + {Tan}^{- 1} \frac{3}{4})$
D
$5 e^{3 x} \sin (4 x - {Tan}^{- 1} \frac{3}{4})$

Solution:

Suppose that $y = e^{3 x} \sin 4 x$

Differentiate both sides

$\begin{array}{rcl} \frac{d y}{d x} & = & e^{3 x} (4 \cos 4 x) + 3 e^{3 x} (\sin 4 x) \\ = & e^{3 x} (3 \sin 4 x + 4 \cos 4 x) \\ = & e^{3 x} (\frac{3}{5} \sin 4 x + \frac{4}{5} \cos 4 x) (d e v i d e w i t h \sqrt{3^{2} + 4^{2}} = 5) \\ = & 5 e^{3 x} (\cos 4 x \sin A + \sin 4 x \cos A) \\ = & 5 e^{3 x} (\sin (A + 4 x)) \end{array}$

Here $\sin A = \frac{4}{5}, \cos A = \frac{3}{5} \Rightarrow \tan A = \frac{4}{3}$

Therefore, $\frac{d y}{d x} = 5 e^{3 x} (\sin (4 x + \tan^{- 1} (\frac{4}{3})))$

The derivative of $e^{3 x} \sin 4 x$ with respect to $x,$ is

Solution:

Related content