Answer:
Step-by-step explanation:
Given the equation 4x²+ 49y² = 196
a) Differentiating implicitly with respect to y, we have;
[tex]8x + 98y\frac{dy}{dx} = 0\\98y\frac{dy}{dx} = -8x\\49y\frac{dy}{dx} = -4x\\\frac{dy}{dx} = \frac{-4x}{49y}[/tex]
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
[tex]y^{2} = \frac{196}{49} - \frac{4x^{2} }{49} \\y = \sqrt{\frac{196}{49} - \frac{4x^{2} }{49} \\} \\[/tex]
Differentiating y with respect to x using the chain rule;
Let [tex]u= \frac{196}{49} - \frac{4x^{2} }{49}[/tex]
[tex]y = \sqrt{u} \\y =u^{1/2} \\[/tex]
[tex]\frac{dy}{dx} = \frac{dy}{du} * \frac{du}{dx}[/tex]
[tex]\frac{dy}{du} = \frac{1}{2}u^{-1/2} \\[/tex]
[tex]\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}\\[/tex]
[tex]\frac{dy}{dx} = \frac{-4x}{7\sqrt{196-4x^{2} } }[/tex]
c) From the solution of the implicit differentiation in (a)
[tex]\frac{dy}{dx} = \frac{-4x}{49y}[/tex]
Substituting [tex]y = \sqrt{\frac{196}{49} - \frac{4x^{2} }{49} \\[/tex] into the equation to confirm the answer of (b) can be shown as follows
[tex]\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}}}[/tex]
This shows that the answer in a and b are consistent.
