Finding derivative of tan^2(sinx)?

Question

I need to find the derivative tan^2(sinx) I know that sinx's derivative is cosx.

Answer 1

You've got to chain the hell out of this.

d/dx[tan^2(sinx)]=

2tan(sinx)sec^2(sinx)cosx

Answer 2

You need the chain rule and the fact that the derivative of tangent is secant squared.
So
dy/dx=
2tan(sinx)sec^2(sinx)(cosx)

Answer 3

Chain rule.

[(tan(sinx))^2]' = 2(tan(sinx))' = 2[sec²(sinx)](sinx)'

= 2[sec²(sinx)](cosx)

Answer 4

=tan^2(x) * sin(x)
= (sin^2(x)/cos^2(x)) *sin(x)
=sin^3(x)/cos(x)

now use division rule or multiplication rule (by saying cos inverse (x)) according to ur convenience

Use chain rule to derive sin^3(x) and derivative of cos(x) is -sin (x)

= 3sin^2(x) cos(x) cos inv (x) + sin^3(x)(-1)(1/cos^2(x)) (-sin (x))
= 3sin^2(x) + ( sin^4(x) / ( cos^2(x) ) )

Answer 5

i love stan's answer. even if, you may want to opt for to rewrite it with the "ln" time period as second, and the organic product time period first. be conscious: you may want to also element out the 4, even if it would want to seem slightly ugly.

Answer 6

ask a math 53 prof.