Question

Derivative of cos^2x
The chain rule gets a different answer than the power reducing formula

Answer 1

`The chain rule gets a different answer than the power reducing formula`
What? 0_o
You must apply both rules, power rule first.

Answer 2

Oh you applying Half-Angle Formula to cosine before differentiating?
Maybe I misunderstood :)

Answer 3

blah I'll wait ^^

Answer 4

If the chain rule is used to find\[\frac{ d }{ dx } \cos^2x = 2\cos(x)(\cos)'=> -2cosx(sinx)\]

Answer 5

If the power reducing formula is used

\[\cos^2x = \frac{ 1 + \cos 2x }{ 2 }\]

\[\frac{ d }{ dx } \frac{ 1 + \cos2x }{ 2 } = -sinx \]

Did I make a mistake somewhere?

Answer 6

\[\large\rm \frac{d}{dx}\frac{1}{2}+\frac{1}{2}\cos(2x)\quad=0+\frac{1}{2}(-\sin(2x))(2x)'\]

Answer 7

\[\large\rm =-\sin(2x)\]Ya? :)

Answer 8

cosine turns into minus sine,
(The inner function shouldn't change at this point, the angle is still 2x)
then chain rule

Answer 9

So it's this,\[\large\rm \frac{d}{dx}\cos(2x)\quad\approx-\sin(2x)\]but with chain rule,\[\large\rm \frac{d}{dx}\cos(2x)\quad=-\sin(2x)(2x)'\]\[\large\rm \frac{d}{dx}\cos(2x)\quad=-2\sin(2x)\]Confusing? :o

Answer 10

Using the power reducing formula y' = -sin(2x)

But using the chain rule for cos^2x it is −2cosx(sinx)

Answer 11

Good observation! :)
Recall your Sine Double-Angle Formula:\[\large\rm \sin(2x)=2sinxcosx\]

Answer 12

Thank you very much!

Answer 13

np