1. ## Chain rule stuff

Fine the derivative of the function:
1. g(x)=(1+4x)^5 (3+x-x^2)^8
2. y(x)=2^(sin(pie)x)
3. (1+cos^2x)^6

i dont get this **** and my test is on thursday.

2. Originally Posted by jeph
Fine the derivative of the function:
1. g(x)=(1+4x)^5 (3+x-x^2)^8
First use the product rule,
[(1+4x)^5]'(3+x-x^2)+(1+4x)^5[(3+x-x^2)^8]'
Now use the chain rule,
[5*4(1+4x)^4](3+x-x^2)+(1+4x)^5[8(1-2x)(3+x-x^2)^7]
Simplify,
20(1+4x)^4(3+x-x^2)+8(1+4x)^5(1-2x)(3+x-x^2)^7
Factor,
(1+4x)^4(3+x-x^2)[20+8(1+4x)(1-2x)(3+x-x^2)^6]

3. Originally Posted by jeph
2. y(x)=2^(sin(pie)x)
We can express 2^x as e^(ln 2 * x)
Thus,
2^[sin (pi x)]=e^[ln 2 *sin (pi x)]
Now, the chain rule says that,
e^{f(x)} then the derivative is, f'(x)e^{f(x)}
That is take derivate of exponent and bring it down.
In this case the exponent is,
ln 2* sin(pi x) where ln 2 is a constant function,
The derivative is after using chain rule,
pi*ln 2*cos(pi x)
(pi*ln 2*cos(pi x))e^[ ln 2*sin (pi x)]
But, the 'e' exponent simplifies to (look at beginning of post),
(pi*ln 2*cos(pi x))2^[sin (pi x)]

4. Originally Posted by jeph
3. (1+cos^2x)^6

.
You use a double chain rule here.
The inner function is,
1+[cos x]^2
The derivative of this function is,
-2 sin x cos x=-sin 2x (double angle)
Thus, derivative of
(1+cos^2 x)^6 is,
Inner function derivative which we found to be,
-sin 2x
Times outer function derivative,
6(1+cos^2 x)^5
Multiply them out,
-6*sin 2x(1+cos^2 x)^5

5. sorry about using that word, i didnt think it would have been considered inappropriate =/

can you show me how
1+[cos x]^2 = -2 sin x cos x

this whole time i tohught i wasnt getting the chain rule, but this is the part that got me.

6. Originally Posted by jeph
sorry about using that word, i didnt think it would have been considered inappropriate =/

can you show me how
1+[cos x]^2 = -2 sin x cos x

this whole time i tohught i wasnt getting the chain rule, but this is the part that got me.

New questions should be posted in new threads!!

Are you asking how to take the derivative of 1 + [cos(x)]^2? Because

1+[cos x]^2 is most certainly NOT -2 sin x cos x. (For starters the LHS is always positive.)

d/dx{1 + [cos(x)]^2} = 0 + d/dx{[cos(x)]^2}

Your "outer function" is y^2 where y = cos(x) and your "inner function" is cos(x), so take the derivative of y^2 (2*y) and multiply that by the derivative of cos(x): ( [f(g(x))]' = f'(g(x))*g'(x) )
d/dx{[cos(x)]^2} = 2*cos(x)* -sin(x) = -2sin(x)cos(x)

-Dan

7. Originally Posted by topsquark
New questions should be posted in new threads!!
I do not consider that to be a violation. Since that question is asking in my solution which the user did not understand.

8. Originally Posted by ThePerfectHacker
I do not consider that to be a violation. Since that question is asking in my solution which the user did not understand.
(Shakes his head in despair) Caught not reading the whole question again wasn't I? I tell you, I'm on a roll.

-Dan