# Thread: Help with 2 complicated chain rule problems

1. ## Help with 2 complicated chain rule problems

1. Suppose that

Find an equation for the tangent line to the graph of at .
Tangent line: = ??????

2. Suppose that

f'(x)= 4(6x+2)^3 (6)..this is what i got.

(A) Find an equation for the tangent line to the graph of at .
Tangent line: = ???
B) Find the value of where the tangent line is horizontal.
= ????

2. Just in case a picture helps...

Straight continuous lines differentiate downwards with respect to x, the straight dashed line with respect to the dashed balloon expression, for the sake of the chain rule. But the chain rule is applied inside of the product rule, highlighted if we enclose some balloons in more...

Hope this helps, or doesn't further confuse.

Don't integrate - balloontegrate!

http://www.ballooncalculus.org/examples

3. Then plug in 2 as your value of x for f(x) and f'(x), to find y = mx + c.

$\displaystyle f(2) = 20$ and $\displaystyle f'(2) = -220$

So,

$\displaystyle y = m(x) + c$

can be

$\displaystyle 20 = -220(2) + c$

and so on.

Your derivative in 2 is fine, so plug in similarly...

4. Originally Posted by Kayla_N
1. Suppose that

Find an equation for the tangent line to the graph of at .
Tangent line: = ??????

2. Suppose that

f'(x)= 4(6x+2)^3 (6)..this is what i got.

(A) Find an equation for the tangent line to the graph of at .
Tangent line: = ???
B) Find the value of where the tangent line is horizontal.
= ????

For the first one you need the quotient rule.
The derivative of $\displaystyle {f\over g}={f'g-fg'\over g^2}$
THEN you need to simplify your work.

Your derivative in 2 is correct.
NOW you need to find m and b where y=mx+b.
In order to do that you need a point (x,y).
They gave you x=2. Thus y=f(2), so just plug in.
And m is the slope at that point, hence $\displaystyle m=f'(2)$.
THEN, the horizontal tangent point is where f'(x)=0.
That's easy, it's when 6x+2=0, which is........

5. quotient rule: (5-3x)^4(10x)-5x^2(4)(5-3x)^3(3) / (5-3x)^5

6. Originally Posted by Kayla_N
quotient rule: (5-3x)^4(10x)-5x^2(4)(5-3x)^3(3) / (5-3x)^5
2 errors...
(1) chain rule -3 not 3.
(2) denominator

$\displaystyle {(5-3x)^4(10x)-5x^2(4)(5-3x)^3(-3) \over (5-3x)^8}$

Next get y=f(2) and m=f'(2)
then from y=mx+b you can obtain b.
Once you have m and b you're done.