What is the derivative of a logarithmic trigonometric function?

[davemoosehead]

I'm not that good with trig problems, I don't know what it is. Here are some answers I got, just wondering if they are right.

1) If x cos (y) + y cos (x) = 1, find an expression for dy/dx

Ok so using the product rule I got
(x)(-sin y)+(1)(cos y) + (y)(-sin x)+(1)(cos x) = 0 =>
-x sin (y) + cos (y) - y sin (x) + cos (x)

2) f(t) = tan(sin t²)

f'(t) = sec² (sin t²)(2sin t)(cos t)
can this be reduced? do I have the brackets right?

3) Find the value of lim x->0 (tan 2x)/x

I plugged sin/cos in for tan and got
( sin 2x/cos 2x ) / (x) =>
( sin 2x/cos 2x ) * (1/x)

but now I'm stuck

4) Find y' if y = log (base 3) (x²e^x)
y' = (x²e^x + 2xe^x) / (x²e^x)(ln 3)
factor out x²e^x
y' = (x+2)/(x ln 3)

 

[mjsd]

I'm not that good with trig problems, I don't know what it is. Here are some answers I got, just wondering if they are right.

1) If x cos (y) + y cos (x) = 1, find an expression for dy/dx

Ok so using the product rule I got
(x)(-sin y)+(1)(cos y) + (y)(-sin x)+(1)(cos x) = 0 =>
-x sin (y) + cos (y) - y sin (x) + cos (x)

is x or y a function of each other?

2) f(t) = tan(sin t²)

f'(t) = sec² (sin t²)(2sin t)(cos t)
can this be reduced? do I have the brackets right?

your f'(t) is wrong

3) Find the value of lim x->0 (tan 2x)/x

I plugged sin/cos in for tan and got
( sin 2x/cos 2x ) / (x) =>
( sin 2x/cos 2x ) * (1/x)

use L'Hopital's rule (ans: 2)

4) Find y' if y = log (base 3) (x²e^x)
y' = (x²e^x + 2xe^x) / (x²e^x)(ln 3)
factor out x²e^x
y' = (x+2)/(x ln 3)

wrong again

EDIT: sorry this one is ok (ignore my previous comment)

 

[davemoosehead]

1) Working...

2) How far off am I? All you do is apply the chain rule right?

3) Is there a way to solve this without l'hospital's rule? I don't think we've learned it yet...

 

[davemoosehead]

1) Since y is a function of x
-xy'sin(y) + cos(y) - ysin(x) + y'cos(x) = 0
y'(-xsin(y)) + y'cos (x) = ysin(x) - cos(y)
y' = (ysin (x) - cos (y)) / (-xsin(y) + cos(x))

 

[mjsd]

2) ... All you do is apply the chain rule right?

yes..you were close

1) Since y is a function of x

use implicit differentiation... and you have got it worked out.

 

[mjsd]

for 3)
note that
$$\tan(z) = z + \frac{z^3}{3} + \frac{2z^5}{15} +\frac{17 z^7}{315} +O(z^8)$$
divide thru and then sub in value...

 

[davemoosehead]

2) f'(t) = sec²(sin t²)(cos t²)(2t)

 

[mjsd]

2) f'(t) = sec²(sin t²)(cos t²)(2t)

yes.

 

[davemoosehead]

Thanks for your help

For 3) I ended up getting

lim x-> 0 (sin x /x)+(sin x /x) / cos 2x = 2

 

[mjsd]

For 3) I ended up getting

lim x-> 0 (sin x /x)+(sin x /x) / cos 2x = 2

here you have used sin x/x ->1 as x->0 which is usually proved using L'Hopital rule...or of course can be seen by expanding into power series