Question

find the derivative
4x + 5 sin x
÷
3x + 7 cos x​

Answer 1

The derivative of $\dfrac{4x+5\sin x}{3x+7\cos x}$ = $\dfrac{(3x+7\cos x)(4+5\cos x)-(4x+5\sin x)(3-7\sin x)}{(3x+7\cos x)^2}$

Step-by-step explanation:

Let $y=\dfrac{4x+5\sin x}{3x+7\cos x}$

To find, the derivative of $\dfrac{4x+5\sin x}{3x+7\cos x}$ = ?

∴ $y=\dfrac{4x+5\sin x}{3x+7\cos x}$

Using division formula,

$\dfrac{x(\dfrac{u}{v})}{dx} =\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{v^2}$

$\dfrac{dy}{dx} =\dfrac{(3x+7\cos x)\dfrac{d(4x+5\sin x)}{dx}-(4x+5\sin x)\dfrac{d(3x+7\cos x)}{dx}}{(3x+7\cos x)^2}$

⇒ $\dfrac{dy}{dx} =\dfrac{(3x+7\cos x)(4+5\cos x)-(4x+5\sin x)(3-7\sin x)}{(3x+7\cos x)^2}$

∴ The derivative of $\dfrac{4x+5\sin x}{3x+7\cos x}$ = $\dfrac{(3x+7\cos x)(4+5\cos x)-(4x+5\sin x)(3-7\sin x)}{(3x+7\cos x)^2}$