Derivatives and Inverse Trigonometry

[ardentmed]

Hey guys,

I have a couple of questions about this problem set I've been working on. I'm doubting some of my answers and I'd appreciate some help.

Question:

For 1a, using inverse trigonometric derivative identities should work, right?

I got y' = 1/sinØ + 1/cosØ and multiplied by the common denominator to get.

y' = (cosØ+sinØ)/sinØcosØ

As for 1b, I used the product rule and simplified to get:

y' = 3^x * ln3*ln(3x) + (3^x)/x + 3/x


Also, for 2, I'm a bit confused. I took the derivative and substituted x = $\pi$/2 to get 0 + $\pi$ = $\pi$

So y'($\pi$/2) = $\pi$ Does that look right?

As for the second derivative, I got
y''=16+ 4$\pi$/3

Thanks in advance.

 

[jacobi1]

For 1a, you didn't use the chain rule correctly. You did f'(g(x)), but didn't multiply by g'(x).
The next two are correct.
But \(\displaystyle y''(\frac{\pi}{6}) \) is wrong. How did you do it? Can you show steps?

 

[ardentmed]

For 1a, you didn't use the chain rule correctly. You did f'(g(x)), but didn't multiply by g'(x).
The next two are correct.
But \(\displaystyle y''(\frac{\pi}{6}) \) is wrong. How did you do it? Can you show steps?

Oh, so there should be a coso and sino in the nominator, right?

In that case, after simplifying, I obtained 1-1=0. Is that correct?

Also, for the last part of question 2, I don't see how I could have gotten it wrong unless f''(x) is wrong. I got f''(x) = 2sec^2(2x) + 2sec^2(2x) + 8xsec(2x)cos(2x)

 

[Dethrone]

Yes, there should be a cos and sin in the numerator. Your second derivative is wrong, can you show your steps?

The first derivative is \(\displaystyle f'(x) = \tan\left({2x}\right) +2x\sec^2\left({2x}\right) \)

 

[jacobi1]

After simplifying, you should obtain a constant...but not zero. Could you show all your steps for the inverse trig derivative?

 

[ardentmed]

Yes, there should be a cos and sin in the numerator. Your second derivative is wrong, can you show your steps?

The first derivative is \(\displaystyle f'(x) = \tan\left({2x}\right) +2x\sec^2\left({2x}\right) \)

I did it again and got (48+16√3*$\pi$)/3

I simply took the derivative of tan(2x)+2xsec^2 (2x) and computed:

f''(x) = 2sec^2(2x) + 2sec^2(2x) + 8xsec^2(2x)tan(2x)

Thus,

f''(x) = 2sec^2($\pi$/3) + 2sec^2($\pi$/3) + 4$\pi$/3 * sec^2($\pi$/3)tan($\pi$/3)

 

[Dethrone]

Yup, this answer right now is correct. (Star)

 

[ardentmed]

After simplifying, you should obtain a constant...but not zero. Could you show all your steps for the inverse trig derivative?

Because:

y = sin^-1(cos$\theta$) + cos^-1(sin$\theta$)

Thus,

y' = 1/(√1-cos^2($\theta$)) * (sin$\theta$) - 1/ (√1-sin^2($\theta$)) * cos$\theta$

Thus, knowing that sin^2($\theta$) + cos^2($\theta$) = 1, we know that:

y' = cos($\theta$)/cos($\theta$) - sin($\theta$)/sin($\theta$)

y' = 0.

Where did I go wrong?

 

[Dethrone]

Your chain rule for \(\displaystyle \sin^{-1}\left({\cos\left({x}\right)}\right)\) is incorrect. What is the derivative of $\cos\left({x}\right)$?

 

[ardentmed]

Your chain rule for \(\displaystyle \sin^{-1}\left({\cos\left({x}\right)}\right)\) is incorrect. What is the derivative of $\cos\left({x}\right)$?

It should be -sinx.

So is the answer -1-1= -2?

 

[Dethrone]

That appears to be correct. Glad you found your mistake :D