Calculus Homework Help: Differentiating Trigonometric and Polynomial Functions

[1calculus1]

Homework Statement

1) If f(x)= sin^4x, then f '(pi/3)
2) Given f(x) = x/tanx, find f '(3pi/4)
3) If f(x) = sinxcosx, then f '(pi/6)
4) Differentiate: f(x) = x^2 + 2tanx
Question that I have answered but not sure if it's really the right answer:
5) Find the equation of the tangent line to the graph of f(x) = (x-1) / (x+1) when x = 1

Homework Equations

Product Rule?

• 
  y= f x g
  F'g + g'f

Chain Rule?

• 
  y=f / g
• ((f'g) - (g'f)) / ((g^2))

Slope?

• y=mx+b

The Attempt at a Solution

1)
f(x) = sin^4x then f '(pi/3)
Derivative of sinx = cosx therefore...
I'll assume that sin^4x has the derivative of cos^4x
Now, plug in the number...
cos^4(pi/3) = 0.065

Is that right?

2)
Given f(x) = x/tanx, find f '(3pi/4)
Chain Rule:
((f'g) - (g'f)) / (g^2)
Therefore...
((1 * tanX) - (? * X)) / ((tanx^2))

Let * be multiplication sign and the ? to be the "I don't know".
So, I got stuck of what the derivative of "tanx". However, what I do know is that:
tanx = sinx / cosx
tanx' = cox / -sinx <-------Is that right?

If yes, then how am I suppose to make my equation by using "cox / -sinx "?
By plugging that in... I get this:

(((1 * tanX) - ((cox / -sinx) * X))) / ((tanx^2))
Then I'm really stuck on that one... I mean, if I do plug in the "pi/3" to the "x" variables then it will just be a mess. Unless that's the only way to get the answer? Or should have I used the product rule instead?


3)
If f(x) = sinxcosx, then f '(pi/6)
Product rule:
y= f x g
F'g + g'f
Therefore...
= (cosx*cosx) + (-sinx*sinx)
= (cos^2x) + (-sin^2x)
= (cos^2(pi/6)) + (-sin^2(pi/6))
= 0.633

-Let * be a multiplication sign
-Is that right?

4)
Differentiate: f(x) = x^2 + 2tanx

So, I'll just get the derivative of the equation...
2x + 2(?)

-Let ? be "I don't know".
So, I'm stuck. I have no idea what's the derivative of tanx have. I already encountered this problem in question #3 and I assumed that it would be:

tanx = sinx / cosx
tanx' = cox / -sinx

Is that right? If yes, then I would get this equation:
2x + 2(cosx/-sinx)

Is that right? If yes, can I simplify it much more?

5)
Find the equation of the tangent line to the graph of f(x) = (x-1) / (x+1) when x = 1
y=mx+b
Chain Rule:
= 1(x+1) - 1(x-1) / (x+1)^2
= x+1 -x +1 / (x+1)^2
= 2 / (x+1)^2
= Plug in "x"
= 2 / (1+1)^2
= 2 / 4
= 1/2
slope (m) = 1/2

Now that I have the slope, I'll just get the x & y values from plugging in "1" to the equation.
y = (1-1) / (1+1)
y = 0/2
y = 0
So: x = 1 and y = 0
Then reflect on the slope equation:
y = mx+b
Plug in the numbers from what I have gotten before:
0 = 1/2(1)+b
0 - 1/2 = b
-1/2 = b

So...
y = 1/2(x) + (-1/2)
I'll multiply the whole equation by "2" to make it more neater.
2y = 2(1/2x) + 2(-1/2)
2y = x -1

Is that right?



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I know my solutions were kind of long. But I hope that you could help me. I really want to make this happen. Or at least answer the questions correctly or in a much simplified way. =)

 

[1calculus1]

Please.

 

[1calculus1]

I've read the "FAQ: Why hasn't anybody answered my question?" but it seems like nobody is answering. Maybe its due to the fact that they don't know what I'm talking about but or they are just too lazy to actually read it all.

But, PLEASE, if you are just lazy...
Please answer this one simply question:

What's the derivative of tan(X)?
Yes, I know that tan(x) = sinx/cosx therefore...
tanx' = cosx/-sinx
?
Is that right?
PLEASE help.
Please.

 

[nicksauce]

It seems like you need to learn, or at least how to correctly apply, all your elementary rules of differentiation."f(x) = sin^4x then f '(pi/3)
Derivative of sinx = cosx therefore...
I'll assume that sin^4x has the derivative of cos^4x"

That is wrong. If f(x) = sin^4x, then f '(x) = 4 * sin^3 x * cosx. You must apply the chain rule here.

"
Yes, I know that tan(x) = sinx/cosx therefore...
tanx' = cosx/-sinx "

That is also wrong. You must apply the quotient rule here. The correct answer is tanx' = sec^2 x.

 

[1calculus1]

It seems like you need to learn, or at least how to correctly apply, all your elementary rules of differentiation.


"f(x) = sin^4x then f '(pi/3)
Derivative of sinx = cosx therefore...
I'll assume that sin^4x has the derivative of cos^4x"

That is wrong. If f(x) = sin^4x, then f '(x) = 4 * sin^3 x * cosx. You must apply the chain rule here.

"
Yes, I know that tan(x) = sinx/cosx therefore...
tanx' = cosx/-sinx "

That is also wrong. You must apply the quotient rule here. The correct answer is tanx' = sec^2 x.

Thanks for the reply. I really appreciate it.
However, I am still having trouble on understand how you got "cosx" from the first question. Yes, the derivative of sinx is cosx but how did you get it to blend in the derivative answer?

Next question...
Alright. So, I applied the quotient rule so and this is what it gave me:
f'g+g'f
cosx/sinx
= (cosx)(cosx) + (-sinx)(sinx)
= cos^2x - sin^2x
= I don't get it. Unless I apply the chain rule which is...
(f'g - g'f) / (g)^2
then I can get this:
= ((cosx)(cosx) - (-sinx)(sinx)) / (sinx)^2
= (cos^2x + sin^2x) / (sinx)^2
= 1 / sinx^2
= cscx^2
which is wrong from your perspective.

Please help, again.

 

[HallsofIvy]

Thanks for the reply. I really appreciate it.
However, I am still having trouble on understand how you got "cosx" from the first question. Yes, the derivative of sinx is cosx but how did you get it to blend in the derivative answer?

Next question...
Alright. So, I applied the quotient rule so and this is what it gave me:
f'g+g'f

NO, that is NOT the "quotient rule", it is the "product rule".

cosx/sinx
= (cosx)(cosx) + (-sinx)(sinx)
= cos^2x - sin^2x
= I don't get it. Unless I apply the chain rule which is...
(f'g - g'f) / (g)^2

NO, that is NOT the "chain rule", that is the "quotient rule".

then I can get this:
= ((cosx)(cosx) - (-sinx)(sinx)) / (sinx)^2
= (cos^2x + sin^2x) / (sinx)^2
= 1 / sinx^2
= cscx^2
which is wrong from your perspective.

Please help, again.

PLEASE read your textbook carefully and learn the correct rules. Especially the chain rule- that is the most helpful.