How do you differentiate sinx+xy+y^5=pi?

Aug 27, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{y + \cos x}{x + 5 {y}^{4}}$

Explanation:

$\sin x + x y + {y}^{5} = \pi$
Differentiating both sides with respect to $x$ we get
$\cos x + \left(y + x \frac{\mathrm{dy}}{\mathrm{dx}}\right) + 5 {y}^{4} \frac{\mathrm{dy}}{\mathrm{dx}} = 0$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{y + \cos x}{x + 5 {y}^{4}}$

Aug 27, 2017

Given: $\sin x + x y + {y}^{5} = \pi$

Differentiate each term:

$\frac{d \left(\sin x\right)}{\mathrm{dx}} + \frac{d \left(x y\right)}{\mathrm{dx}} + \frac{d \left({y}^{5}\right)}{\mathrm{dx}} = \frac{d \left(\pi\right)}{\mathrm{dx}}$

The first term is well known $\frac{d \left(\sin x\right)}{\mathrm{dx}} = \cos \left(x\right)$:

$\cos \left(x\right) + \frac{d \left(x y\right)}{\mathrm{dx}} + \frac{d \left({y}^{5}\right)}{\mathrm{dx}} = \frac{d \left(\pi\right)}{\mathrm{dx}}$

The second term requires the use of the product rule:

$\frac{d \left(x y\right)}{\mathrm{dx}} = \frac{d \left(x\right)}{\mathrm{dx}} y + x \frac{d \left(y\right)}{\mathrm{dx}}$

$\frac{d \left(x y\right)}{\mathrm{dx}} = y + x \frac{\mathrm{dy}}{\mathrm{dx}}$

$\cos \left(x\right) + y + x \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{d \left({y}^{5}\right)}{\mathrm{dx}} = \frac{d \left(\pi\right)}{\mathrm{dx}}$

The third term requires the use of the chain rule:

$\frac{d \left({y}^{5}\right)}{\mathrm{dx}} = \frac{d \left({y}^{5}\right)}{\mathrm{dy}} \frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{d \left({y}^{5}\right)}{\mathrm{dx}} = 5 {y}^{4} \frac{\mathrm{dy}}{\mathrm{dx}}$

$\cos \left(x\right) + y + x \frac{\mathrm{dy}}{\mathrm{dx}} + 5 {y}^{4} \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d \left(\pi\right)}{\mathrm{dx}}$

For the last term, we invoke the fact that the derivative of a constant is 0:

$\cos \left(x\right) + y + x \frac{\mathrm{dy}}{\mathrm{dx}} + 5 {y}^{4} \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

Move all of the terms NOT containing $\frac{\mathrm{dy}}{\mathrm{dx}}$ to the right:

$x \frac{\mathrm{dy}}{\mathrm{dx}} + 5 {y}^{4} \frac{\mathrm{dy}}{\mathrm{dx}} = - \left(\cos \left(x\right) + y\right)$

Factor out $\frac{\mathrm{dy}}{\mathrm{dx}}$ on the left:

$\left(x + 5 {y}^{4}\right) \frac{\mathrm{dy}}{\mathrm{dx}} = - \left(\cos \left(x\right) + y\right)$

Divide both sides by the leading coefficient:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{\cos \left(x\right) + y}{x + 5 {y}^{4}}$