# Thread: Chain Rule, and Power Rule

1. ## Chain Rule, and Power Rule

Can anyone help me out... I know that on problem like $\displaystyle 5(5+5x^2)^5 y= 5u^5 , u=5+5x^2$ you find the derivtive of those using simple differentiation, and them combine and so on.

However, I'm not sure how to do this: find the equation of a tangent line to the graph of f using x=-2 on an equation of $\displaystyle f(x)=6x^2(3-6x^2)^5$

I thought that was the derivative, how can this be accomplished. Simply plugging in -2 did not work. .

2. I'm confused a bit by your wording. Are you saying f(x)=6x^2..... and you need to find the tangent line to f(x) at x=-2?

3. Sorry, I should of said: Find an equation for the tangent line to the graph f at x=-2

4. Ok, so $\displaystyle f(x)=6x^2(3-6x^2)^5$

Plugging in x=-2 gives you the y-value of this point on the function. So you'll have a point (-2, f(-2)).

Now take the derivative of f(x) and plug in -2 to find the slope of the tangent line at x=-2.

Now you have a point and a slope, thus a line can be formed!

Point-slope formula: $\displaystyle y-y_0=m(x-x_0)$

$\displaystyle (x_0 , y_0)$ is the point on the graph you found and m is the slope at x=-2.

Make sense?

5. How can the derivitive of this function be solved using the chain or power rule?

6. Ok so $\displaystyle f(x)=6x^2(3-6x^2)^5$. We have a product here, thus the product rule, and the second term is a composite expression thus the chain rule is needed.

$\displaystyle f'(x) = 6x^2 \times \frac{d}{dx} \left( 3-6x^2 \right)^5 + (3-6x^2)^5 \times \frac{d}{dx} \left( 6x^2 \right)$

$\displaystyle f'(x)=6x^2[5(3-6x^2)^4 \times (-12x)] + (3-6x^2)^5 \times (12x)$.