Question

Derivative of trigonometric functions
1.) y=sin2x cos2x
2.) y=cos^4x - sin^4x
3.) y=cotx - tanx

Answer 1

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Answer 2

well if it makes the first one easier you use an identity first
\[2 \sin(u)\cos(u)=\sin(2u) \\ \sin(u)\cos(u)=\frac{1}{2} \sin(2u) \\ \text{ so you can say you have } \\ \sin(2x)\cos(2x)=\frac{1}{2} \sin(2 \cdot 2x)=\frac{1}{2} \sin(4x)\]
so for the first one you are really looking at differentiating
1/2*sin(4x)

Answer 3

so you get the avoid the product rule for the first one

Answer 4

apply constant multiple rule and chain rule

Answer 5

for the second one you could use the identity
cos2x = cos^2x - sin^2x

Answer 6

@pearlss are you there?

Answer 7

like do you have any thoughts on what @welshfella or I said? like let me know if you don't understand something?

Answer 8

Derivative of trigonometric functions

3.) y=cotx - tanx

Answer 9

have you already found the derivative of the first two?

Answer 10

number 3 is probably the easiest
since we have
\[\frac{d}{dx}\cot(x)=-\csc^2(x) \\ \text{ and } \\ \frac{d}{dx}\tan(x)=\sec^2(x)\]

Answer 11

the other two take a little more work

Answer 12

@freckles the derivative of no. 2 i need the solution

Answer 13

@welshfella gave a pretty hint for number 2

Answer 14

\[\cos^4(x)-\sin^4(x) \\ (\cos^2(x)-\sin^2(x)) \cdot (\cos^2(x)+\sin^2(x)) \text{ since we have a difference of squares } \\ \cos(2x) \cdot (1) \\ \text{ by double \angle identity for cosine and by pythagorean identity } \\ \cos^4(x)-\sin^4(x)=\cos(2x) \\ \text{ now you just need \to evaluate } \frac{d}{dx} \cos(2x)\]

Answer 15

you can apply chain rule

Answer 16

@ freckles and welshfella Thank you very much ^_^

Answer 17

I guess you have no problem finding the derivatives of the prettier functions we gave?