Calculus help pleaseeeee :(

[i Maax i]

Find dy/dx if y=(x^2+5)^10

 

[Koot]

Did you learn product rule? That's what you need to find the derivative of sinxcosx. I'll paste over what I put in the other guys thread.

Product rule is

y = uv
y' = u'v + uv'

Say y = xlnx

y' = x 1/x + (1)lnx
y' = lnx + 1


Now that you have an example, try it again and I'll see if it is correct.

For F'(x), and you want to say find when x=0, just plug it in. F'(0) = .... whereever there are x's plug in 0 and solve it. You have the 2nd part right there but the derivative isn't. Just to answer it, sin pi/3 cos pi/3 is solvable, you can draw a triangle out to demonstrate it. Remember pi = 180 degrees so pi/3 is 60. Sin 60 is rt3 / 2.

 

[i Maax i]

Yeah I learned the product rule. The pi/3 is messing me up.

How'd you know that sin60 is √3/2 ?

Here's what I found:

F'(pi/3) = F'(sin(pi/3) cos(pi/3)) = (cos(pi/3))(-sin(pi/3)) = (√3/2)(-1/2)

Is that good?

 

[video]

You didn't do the product rule correctly. You should have gotten F'(x)=sin(x)*-sin(x)+cos(x)*cos(x). Look at koot's post for clarification on how I got this.

The sin of pi/3=sqrt(3)/2, and the cos of pi/3=1/2. It's just something that you memorize.

Take a look at this for help:

Your answer should be -1/2

 

[i Maax i]

Yeah I just printed something similar and my book says -1/2 too.

I'm going to try again I think I know where I messed up; I forgot to do the product rule before derivating my fontion.

 

[DtheK]

So... if:
F(x)= sinx cosx

Find:
F'(x)
F'(pi/3)

Second:
F'(pi/3)= F'(sin(pi/3) cos(pi/3)) = I'm stuck there.

First:

F(x)=sinxcosx
F'(x)= -(sinx)^2+(cosx)^2
F'(pi/3)= -(sqrt(3)/2)^2+(1/2)^2 = -(3/4)+(1/4) = -2/4 = -1/2

 

[i Maax i]

First:

F(x)=sinxcosx
F'(x)= -(sinx)^2+(cosx)^2
F'(pi/3)= -(sqrt(3)/2)^2+(1/2)^2 = -(3/4)+(1/4) = -2/4 = -1/2

Ya thanks, I finally arrived to the answer after realizing I forgot to put my god damn calculator in RAD mode..........................................