Check Work - Differentiating Trig Function

[JSmith]

1. Differentiate the function f(x)=cos³xsinx.

f'(x)=[(-3sin²x)(sinx)] + [(cos³x)(cosx)]
f'(x)=-3sin³x + cos⁴x

2. Differentiate the function f(x)=sin(tanx).

f'(x)= cos [(tanx)(sec²x)]

3. Differentiate the function f(x)=sin(cos(x²).

f'(x)= cos(cos(x)) (-sin(x²)) (2x)
f'(x)= cos(cos(x)) (-2xsin(x²))

I was wondering if these were correct? They were some of the more tricky ones in my homework. Thanks!!

 

[DrPhil]

1. Differentiate the function f(x)=cos³xsinx.

f'(x)=[(-3sin²x)(sinx)] + [(cos³x)(cosx)] X
f'(x)=-3sin³x + cos⁴x

2. Differentiate the function f(x)=sin(tanx).

f'(x)= cos [(tanx)(sec²x)] not quite

3. Differentiate the function f(x)=sin(cos(x²).

f'(x)= cos(cos(x)) (-sin(x²)) (2x) close
f'(x)= cos(cos(x)) (-2xsin(x²))

I was wondering if these were correct? They were some of the more tricky ones in my homework. Thanks!!

YES, these are very tricky!

1. When you differentiate the cos^3(x) factor, you need to apply the power law first:
d/dx(cos^3x) = 3 cos^2(x) (-sinx)

2. When you use the chain rule, first you get cos(tanx) and then you MULTIPLY by the derivative of tanx. Remove the "[..]" because the sec^2 is not part of the argument of the cosine.

3.You are missing the square of x in the arguments of the cosine.
cos(cos(x^2)) (-sin(x^2)) (2x)

 

[JSmith]

YES, these are very tricky!

1. When you differentiate the cos^3(x) factor, you need to apply the power law first:
d/dx(cos^3x) = 3 cos^2(x) (-sinx)

2. When you use the chain rule, first you get cos(tanx) and then you MULTIPLY by the derivative of tanx. Remove the "[..]" because the sec^2 is not part of the argument of the cosine.

3.You are missing the square of x in the arguments of the cosine.
cos(cos(x^2)) (-sin(x^2)) (2x)

Ok, thanks. I think I follow, but I am not totally sure. Would you mind posting the correct derivative for each then?

 

[soroban]

Hello, JSmith!

Sorry . . . all your answers are wrong.

\(\displaystyle \text{1. Differentiate: }\:f(x) \:=\:\cos^3\!x\sin x\)

\(\displaystyle \text{We have: }\:f(x) \;=\;\underbrace{\cos^3\!x}_{g(x)}\cdot \underbrace{\sin x}_{h(x)} \)

\(\displaystyle \text{Product Rule:}\)

. . \(\displaystyle f'(x) \;=\;\underbrace{\cos^3\!x}_{g(x)}\cdot \underbrace{\cos x}_{h'(x)} + \underbrace{3\cos^2\!x(\text{-}\sin x)}_{g'(x)}\cdot\underbrace{\sin x}_{h(x)}\)

. . \(\displaystyle f'(x) \;=\;\cos^4\!x - 3\cos^2\!x\sin^2\!x\)

\(\displaystyle \text{2. Differentiate: }\:f(x) \:=\:\sin(\tan x)\)

Chain Rule: . \(\displaystyle f'(x)\:=\:\cos(\tan x)\cdot\sec^2\!x\)

\(\displaystyle \text{3. Differentiate: }\:f(x)\:=\:\sin\left[\cos(x^2)\right]\)

\(\displaystyle \text{Chain Rule: }\:f'(x) \;=\;\cos\left[\cos (x^2)\right] \cdot\left[\text{-}\sin(x^2)\right]\cdot 2x\)

n . . . . . . . .. \(\displaystyle f'(x) \;=\;-2x\cdot\cos\left[\cos(x^2)\right]\cdot\sin(x^2)\)