Question

Help please!
can someone show me the steps of how to differentiate the following with respect to x

3sin^2x+sec2x

Answer 1

ddxsin(ax)dx=acos(ax) use this along with chain rule for 3sin^2x

Answer 2

differentiate each term in turn
for 3 sin^2x use the chain rule f'(x) = f'(g(x) * g'(x)
also use chain rule for second term

Answer 3

d/dx [3sin^2(x)] = 6sin x cos x

Answer 4

yup

Answer 5

Let y=sec 2x
dy/dx=d{sec 2x}/dx
=sec 2x tan2x×d(2x)/dx USing chain rule
=sec 2x tan2x×(2)
=2sec 2x tan2x

Answer 6

@jetly
did you use the chain rule to get 6sin x cos x?

Answer 7

yes kinzan.

Answer 8

are you happy with the f'(x) notation or do you usually use the Leibnitz fom r kinsan : - dy/dx etc

Answer 9

*Leibnitz form

Answer 10

i use the the f'(x) notation but do not get how jetly go 2sec 2x tan2x as his final answer...
arnt you meant to use a constant 'u' in the chanirule?

Answer 11

you can use 'u' for the 'inside function' when using the chain rule - some people find it easier to understand

Answer 12

i'll differentiate for you using 'u' and leibnitz notation;

let y = sec2x
and let u = 2x
so y = secu
dy/du = secu tanu
du/dx = 2
so dy/dx = dy/du * du/dx = 2 secu tan u = 2 sec2x tan 2x