Question

Check that your solutions to part (a) and (b) are consistent by substituting the expression for y into your solution for part (a).
9x^2 - y^2 = 1

Answer 1

Answer:

Step-by-step explanation:

The question is incomplete. Find the complete question in the attached file.

a) Given the expression 9x^2 - y^2 = 1

Differentiating implicitly will give;

$18x-2y\frac{dy}{dx} = 0$

We can then make dy/dx the subject of the formula as shown;

$18x = 2y\frac{dy}{dx}\\\frac{dy}{dx} = \frac{18x}{2y} \\\frac{dy}{dx} = \frac{9x}{y} \\y' = \frac{9x}{y}$

b) In order to solve the question explicitly, we will first have to make x the subject of the formula before differentiating.

$9x^{2} -y^{2} = 1\\9x^{2} = 1+ y^{2}\\y^{2} =9x^{2}- 1\\y^{2} = 9x^{2} - 1\\y = \sqrt{9x^{2}- 1 }$

Using chain rule to solve the equation;

let;

$u = 9x^{2} -1\\ y = u^{1/2}$  

du/dx = 18x

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

dy/dx = dy/du * du/dx

dy/dx = $1/2u^{-1/2} * 18x$

$\frac{dy}{dx} = \frac{1}{2}({9x^{2}-1 } )^{-1/2} * 18x\\\frac{dy}{dx} = 9x({9x^{2} -1 } )^{-1/2} \\\frac{dy}{dx} = 9 x(\sqrt{{\frac{1}{9x^{2} -1} } }) \\\frac{dy}{dx} = \frac{9x}{\sqrt{9x^{2}-1 } }$

c) In order to confrim that solutions to part (a) and (b) are consistent, we will substitute $y = \sqrt{9x^{2} - 1 }$ into the answer in (a) as shown;

From (a) $\frac{dy}{dx} = \frac{9x}{y}$

$y' = \frac{9x}{ \sqrt{9x^{2} - 1 } \\} } \\}$

This shows that they are consistent