# Implicit differentiation, equation of the tangent line

Please see the attached problem for the full questions.

1) Find f'(x) and f'(c): f(x) = x^2 - 4 / x - 3, c =1

2) Find dy/dx by implicit differentiation: sinx + 2cos2y = 1

3)

A) Use implicit differentiation to find an equation of the tangent line to the ellipse:

x^2/2 + y^2/8 = 1 at (1,2)

B) Show that the equation of the tangent line to the ellipse 2^2/a^2 + y^2/b^2 = 1 at (x_0,y_0) is

x_0 x/ a^2 + y_0 y/b2^2 = 1

https://brainmass.com/math/derivatives/implicit-differentiation-equation-tangent-line-493852

## SOLUTION This solution is **FREE** courtesy of BrainMass!

6. Find and :

Applying the quotient rule and chain rule

Then substitute 1 for x to find out f'(x=1)

9. Find by implicit differentiation:

Applying chain rule, and differentiating on both sides of equation:

So

10. (a) Use implicit differentiation to find an equation of the tangent line to the ellipse .

Based the definition, the slope of the tangent line is

First find the general formula of the slope of the tangent line:

Differentiate with respect to x,

Then

That is, the slope of the tangent line is

At point (1, 2), the slope of the tangent line is

Now we just need to find the line equation with slope and passing through point (1, 2). Based on the definition of slope, we have

That is, the tangent line to the ellipse at point (1, 2) is

(b) Show that the equation of the tangent line to the ellipse at is .

First find the slope of the tangent line by differentiating both sides of the ellipse equation:

So the slope of the tangent line at point is

Then the line equation with the above specific slope and passing through point is:

Dividing from both sides of equation:

Since point is on the ellipse, it then satisfies the equation .

Hence, the tangent line equation can be written as:

https://brainmass.com/math/derivatives/implicit-differentiation-equation-tangent-line-493852