Question

Find the derivative of the function f(x) = cos^2(x) + cot^2(x).

2cos(x) + 2cot(x)

2cos(x)sin(x) + 2cot^2(x)

–2sin(x)cos(x) – 2cot(x)csc^2(x)

2cos(x)sin(x) – csc^2(x)

Answer 1

@Michele_Laino

Answer 2

what is the first derivative of
\[{\left( {\cos x} \right)^2}\]

Answer 3

-sin^2(x)

Answer 4

No treat the 2 as a power rule

Answer 5

You see Michele put the 2 in brackets which is the same thing as cos^2x just easier to see what's going on :), this required power + chain rule.

Answer 6

got it :)

Answer 7

so it would be: (-sinx)^2

Answer 8

no I don't think, please keep in mind that
\[{\left( {\cos x} \right)^2} = \cos x \cdot \cos x\]
then apply the product rule

Answer 9

so its -sinx times -sinx

Answer 10

?:)

Answer 11

product rule is this:
\[\frac{d}{{dx}}\left( {fg} \right) = \frac{{df}}{{dx}} \cdot g + f \cdot \frac{{dg}}{{dx}}\]

Answer 12

\[f(x) = \cos^2x \implies~~f'(x) = 2(cosx)*cosx'\]

Answer 13

Does that help? :P

Answer 14

yes! (: i think the answer is
c) –2sin(x)cos(x) – 2cot(x)csc2(x)

Answer 15

hint, if I apply the product rule, I can write:
\[\frac{d}{{dx}}{\left( {\cos x} \right)^2} = \frac{{d\cos x}}{{dx}} \cdot \cos x + \cos x \cdot \frac{{d\cos x}}{{dx}} = ...?\]

Answer 16

You got it math!

Answer 17

thank u!!! :)

Answer 18

thank you!!

Answer 19

I think you may have been confused by the notation by Michele, even though that is the "legit" way, but I use primes at the beginning since it's easier to see what's going on when people first start calculus at least :P