Question

Differentiate the following with respect to x:

f(x)= cos^2(x^2) - x - sin^3(x^3)

Answer 1

\[\cos^2(x^2)-x-\sin^3(x^3)\]
this is it if this is easier to look at.

Answer 2

Here set it up as such, so you can see clearly how to apply the differentiation rules \[f(x) = \left( \cos(x^2) \right)^2-x-\left( \sin(x^3) \right)^3\] should make your life easier now to differentiate.

Answer 3

Notice what the exponents really mean

Answer 4

So I pull them in front?
\[f(x)= 2(cosx^2)-x-3(sinx^3)^2\]

Answer 5

Yes, exactly we use the power rule, but you still have to use chain rule!

Answer 6

oh shoot

Answer 7

So, \[f'(x) = 2\cos(x^2) \times (\cos(x^2))' -1-3(\sin(x^3))^2 \times (\sin(x^3))'\]

Answer 8

what is the derivative of cosx^2? I'm not sure what happens with the x^2. I know cos becomes -sin...

Answer 9

You have to use chain rule again for those :P

Answer 10

please tell me it becomes -2sinx

Answer 11

This seems pretty long. Next step is probably to simplify it...

\[f'(X) = 2(\cos^2)(-2sinx)-1-3(sinx^3)^2(3cosx^2)\]

Answer 12

\[f'(x) = \color{blue}{2\cos(x^2) \times - \sin(x^2) \times (x^2)'} -1-\color{red}{3\sin(x^3))^2 \times \cos(x^3) \times (x^3)'}\]

Answer 13

oh god

Answer 14

I mean the derivatives shouldn't be too bad, just make sure you do it step by step, so you don't make mistakes.

Answer 15

you don't pull the 2 in front of the sin I guess

Answer 16

Lol this is a test question that I have to correct because I scored a C #oops

Answer 17

It should be -2xsin(x^2)

Answer 18

You're multiplying so it doesn't matter

Answer 19

This is old material so i'm having to relearn some of it. Just finished studying MTV and Rolle's and L'Hospital and optimization.

Answer 20

Ah, good ole calc 1

Answer 21

So was the previous equation I had written down correct?

Answer 22

Well the class I'm taking now is AP Calc AB. I'm still in High school

Answer 23

Differentiation is very important in all of those haha, so it's crucial you understand it, it'll get simpler as you practice

Answer 24

lol yeah I get pretty lost with all this. I didn't retain anything from precalculus so that didn't help

Answer 25

Not really :P you didn't take the chain rule the second time and you forgot to include the x^2 and x^3 terms in your terms

Answer 26

It's ok, I never took pre calc and jumped into calculus and got wrecked

Answer 27

Took me a while to get it right, so don't get too worried

Answer 28

okay so now I have \[f'(x)= 2(cosx^2)(-sinx^2)(2x)-1-3(sinx^3)^2(cosx^3)(3x^2)\]

Answer 29

looks good!

Answer 30

Now I must struggle through simplifying this if possible

Answer 31

Haha have fun, some of these may come in handy http://www.purplemath.com/modules/idents.htm

Answer 32

Ah I wasn't sure what to combine besides the 2 and the 2x in the first part of the function

Answer 33

Thanks, but I already have all of those memorized. I just struggle with applying them.

Answer 34

Well keep practicing then check with http://www.wolframalpha.com/

Answer 35

That is a beautiful math tool. Thank you.

Answer 36

Yup does wonders :)