Solving Trig Derivatives: Answers & Solutions

[cd246]

Homework Statement

1. Sin^2(t) The answer states sin2t
2. (e^2x)(sin x-cos x) The answer states (e^2x)(3 sin x- cos x)
3. sec^2(x)-tan^2(x)+cos(x) The answer states -sin x

Homework Equations

1.product rule
2.product rule
3. sum and diff. rule?

The Attempt at a Solution

1. (sin t)(sin t)'+(sin t)(sin t)'=(sin t)(cos t)+(sin t)(cos t)=2cos (t) sin (t)
2. (e^2x)(cos x+sin x)+ (e^2x)(sin x-cos x)=(e^2x)(sin 2x)
3. (sec^2(x)*tan^2(x))-sec^4(x)-sin x. I just don't know how to get rid of the first and second functions.

 

[cristo]

Homework Statement

1. Sin^2(t) The answer states sin2t
2. (e^2x)(sin x-cos x) The answer states (e^2x)(3 sin x- cos x)
3. sec^2(x)-tan^2(x)+cos(x) The answer states -sin x

Homework Equations

1.product rule
2.product rule
3. sum and diff. rule?

The Attempt at a Solution

1. (sin t)(sin t)'+(sin t)(sin t)'=(sin t)(cos t)+(sin t)(cos t)=2cos (t) sin (t)

This is correct; you need to use a trig identity to get it in the form of the solution. Note that, alternatively, you could have used the chain rule for this question.

2. (e^2x)(cos x+sin x)+ (e^2x)(sin x-cos x)=(e^2x)(sin 2x)

This isn't quite correct; what is the derivative of e^(2x)?

3. (sec^2(x)*tan^2(x))-sec^4(x)-sin x. I just don't know how to get rid of the first and second functions.

I'm not sure what you've done here. You want to differentiate the original term by term, so you need to find the derivatives of sec^2x and tan^2x. Can you write these as quotients and use the quotient rule?

 

[cd246]

For 1. I tried the chain rule(sin x inside, square outside), and it got me 2cos(t) which is worse than my first answer.
For 2. I assumed e never changes. And I never came across a problem where e changes and I am not too sure on how to do that. Is it 2 e^1?
For 3. If I am understanding this correctly, Instead of the original, say it as: sec^2(x)+cos(x)/tan^2(x) then apply the quotient rule?
If I am not correct, please guide me to the answer(s).

 

[cristo]

For 1. I tried the chain rule(sin x inside, square outside), and it got me 2cos(t) which is worse than my first answer.

You made a mistake using the chain rule. Recall that $$\frac{d}{dx}((\sin x)^2)=2\sin x \cdot\frac{d}{dx}(\sin x)=2\sin x\cos x$$. Now, can you spot the trig identity that will reduce the answer to the required form?

For 2. I assumed e never changes. And I never came across a problem where e changes and I am not too sure on how to do that. Is it 2 e^1?

$$\frac{d}{dx}(e^{ax})=ae^{ax}$$ for some constant a.

For 3. If I am understanding this correctly, Instead of the original, say it as: sec^2(x)+cos(x)/tan^2(x) then apply the quotient rule?
If I am not correct, please guide me to the answer(s).

No, I meant write the expression as $$\frac{1}{\cos^2x}-\frac{\sin^2x}{\cos^2x}+\cos x$$ and differentiate each term, using the quotient rule to differentiate the first two terms.

 

[cd246]

For 1, I am really not seeing how it becomes sin2t from 2sinx cosx.
For 2, I got it right, it was the derivative of e^2x that messed me up. thanks cristo.
For 3, I know ultimately the third term will be left and the other two will cancel each other out. This is what i got(by following the advice):
secx^2-tan^2+cosx
l
\/
(1/cosx^2)-(sinx^2/cosx^2)+cosx.
f'(x)=(cosx^2-sinx^2/(cosx^2)^2)-(cosx^4+sinx^4)/(cosx^2)^2)-sinx. I know the exponents does not cancel in this step.

For 1 and 3, I know there is a step that I do not have any knowledge of(or i did a step wrong), thanks in advance.

 

[cristo]

For 1, I am really not seeing how it becomes sin2t from 2sinx cosx.

Ok, it's just a trig identity you should remember. The identity is $\sin(2x)=2\sin x\cos x$. It is derived from the multiple angle formula $\sin(a+b)=\sin a\cos b+\cos a\sin b$ by setting a=b

For 2, I got it right, it was the derivative of e^2x that messed me up. thanks cristo.

You're welcome

For 3, I know ultimately the third term will be left and the other two will cancel each other out. This is what i got(by following the advice):
secx^2-tan^2+cosx
l
\/
(1/cosx^2)-(sinx^2/cosx^2)+cosx.
f'(x)=(cosx^2-sinx^2/(cosx^2)^2)-(cosx^4+sinx^4)/(cosx^2)^2)-sinx. I know the exponents does not cancel in this step.

Ok, let's look at each term individually, since you have used the quotient rule incorrectly on (at least) the first term.

We want $$\frac{d}{dx}\left(\frac{1}{\cos^2x}\right)$$. Using the quotient rule, this is equal to $$\frac{\cos^2x\cdot\frac{d}{dx}(1)-1\cdot\frac{d}{dx}(\cos^2x)}{(\cos^2x)^2}= \frac{2\cos x\sin x}{\cos^4x}=\frac{2\sin x}{\cos^3x}$$

Now, try to do the second term (sin^2x/cos^2x) again. I'm not sure how you got what you did first time.

 

[cd246]

3. sinx^2/cosx^2
(cosx^2)(sinx^2)'-(cosx^2)'(sinx^2)
------------------------------------
(cosx^2)^2

(cosx^2)(2 cos x sin x)+(2 cos x sin x)(sinx^2)
----------------------------------------------
(cosx^2)^2

(2 cosx^2*cosx^3*sin x cosx^2)+(2sinx^2*cosx sinx^2*sinx^2) distribution
--------------------------------------------------------------
(cosx^2)^2
Is this right so far?

 

[ChaoticLlama]

For #3,

sec^2(x) - tan^2(x) = 1.

the rest is easy.

 

[cristo]

For #3,

sec^2(x) - tan^2(x) = 1.

the rest is easy.

That is indeed a quicker way of doing the question (I can't believe I didn't notice that!) cd246, this comes about from the identity sin^2x+cos^2x=1, and greatly simplifies the work needed.

 

[cd246]

I think I get what u 2 are saying... since sec^2-tan^2 equals 1, it would be 1+cosx, take the derivative and I get -sinx?

 

[cristo]

I think I get what u 2 are saying... since sec^2-tan^2 equals 1, it would be 1+cosx, take the derivative and I get -sinx?

Yup, that's correct.