Question

You are given the following equation.

Answer 1

Answer:

Step-by-step explanation:

Given the equation  4x²+ 49y² = 196

a) Differentiating implicitly with respect to y, we have;

$8x + 98y\frac{dy}{dx} = 0\\98y\frac{dy}{dx} = -8x\\49y\frac{dy}{dx} = -4x\\\frac{dy}{dx} = \frac{-4x}{49y}$

b)  To solve the equation explicitly for y and differentiate to get dy/dx in terms of x,

First let is make y the subject of the formula from the equation;

If 4x²+ 49y² = 196

49y² = 196 - 4x²

$y^{2} = \frac{196}{49} - \frac{4x^{2} }{49} \\y = \sqrt{\frac{196}{49} - \frac{4x^{2} }{49} \\}$

Differentiating y with respect to x using the chain rule;

Let $u= \frac{196}{49} - \frac{4x^{2} }{49}$

$y = \sqrt{u} \\y =u^{1/2}$

$\frac{dy}{dx} = \frac{dy}{du} * \frac{du}{dx}$

$\frac{dy}{du} = \frac{1}{2}u^{-1/2}$

$\frac{du}{dx} = 0 - \frac{8x}{49} \\\frac{du}{dx} =\frac{-8x}{49} \\\frac{dy}{dx} = \frac{1}{2} ( \frac{196}{49} - \frac{4x^{2} }{49})^{-1/2} * \frac{-8x}{49}\\\frac{dy}{dx} = \frac{1}{2} ( \frac{196-4x^{2} }{49})^{-1/2} * \frac{-8x}{49}\\\frac{dy}{dx} = \frac{1}{2} ( \sqrt{ \frac{49}{196-4x^{2} })} * \frac{-8x}{49}\\\frac{dy}{dx} = \frac{1}{2} *{ \frac{7}\sqrt {196-4x^{2} }} * \frac{-8x}{49}$

$\frac{dy}{dx} = \frac{-4x}{7\sqrt{196-4x^{2} } }$

c) From the solution of the implicit differentiation in (a)

$\frac{dy}{dx} = \frac{-4x}{49y}$

Substituting $y = \sqrt{\frac{196}{49} - \frac{4x^{2} }{49}$ into the equation to confirm the answer of (b) can be shown as follows

$\frac{dy}{dx} = \frac{-4x}{49\sqrt{\frac{196-4x^{2} }{49} } }\\\frac{dy}{dx} = \frac{-4x}{49\sqrt{196-4x^{2}}/7} }\\\\\frac{dy}{dx} = \frac{-4x}{7\sqrt{196-4x^{2}}}$

This shows that the answer in a and b are consistent.