Question

find derivative of 5^x sinx​

Answer 1

Given

$\sf \to \: \dfrac{d( {5}^{x} sinx)}{dx}$

Now Using UV method

$\sf \to \: \dfrac{d(uv)}{dx} = v \dfrac{du}{dx} + u \dfrac{dv}{dx}$

Using this Method

$\sf \to \: \dfrac{d( {5}^{x} sinx)}{dx} = {5}^{x} \dfrac{d(sinx)}{dx} + sinx \dfrac{d(5 {}^{x}) }{dx}$

$\sf \to \: \dfrac{d( {5}^{x} sinx)}{dx} = {5}^{x} \times cosx + sinx \times (x {5}^{x - 1} )$

$\sf \to \: \dfrac{d( {5}^{x} sinx)}{dx} = {5}^{x} cosx + x {5}^{x - 1} sinx$

Answer

$\sf \to \: \dfrac{d( {5}^{x} sinx)}{dx} = {5}^{x} cosx + x {5}^{x - 1} sinx$

More Formula

$\to \sf \dfrac{d( {x}^{n}) }{dx} = nx {}^{n - 1}$

$\to \sf \dfrac{d(logx) }{dx} = \dfrac{1}{x}$

$\to \sf \dfrac{d( a) }{dx} = 0$

$\to \sf \dfrac{d( - cosx) }{dx} =sinx$

$\to \sf \dfrac{d( tanx) }{dx} = sec {}^{2} x$

$\to \sf \dfrac{d( - cotx) }{dx} = cosec {}^{2} x$

$\to \sf \dfrac{d( secx) }{dx} = secx \:tanx$

Answer 2

Answer in attachment....