Derivatives of exponent x with a product

[cal.queen92]

Homework Statement

f(x) = 10(sin(x))^x ----> find f '(1)

The Attempt at a Solution

I have tried several different approaches, but still get stuck with a wrong answer every time

f(x) = 10(sin(x))^x let f(x) = y so y=10(sin(x))^x then ln y = ln10(sin(x))^x

using log. laws: lny = xln10(sin(x))

then differentiating implicitly using product and chain rule:

1/y*dy/dx = ln10(sin(x)) + 1/10(sinx) * 10(cosx)*x so

1/y*dy/dx = (ln10(sinx)) + (x(cosx)/(sinx))

then multiplying both sides by y to eliminates denominator:

dy/dx = ((10(sinx))^x)*((ln10(sinx))+(x(cosx)/(sinx))

^ this would be the unsimplified derivative ^

Now for f '(1) I would just plug in 1 where ever there is an x right?

I am doing something wrong, can anyone see my mistake? Thank you!

 

[hgfalling]

\log (10 (\sin x)^x) \neq x \log (10 \sin x)
\log (10^x (\sin x)^x) = x \log (10 \sin x)

 

[cal.queen92]

Hmmm I see, so then what if i try a different approach:

lny = ln10 + xln(sinx)

I continued this way but got the wrong answer again. Is this a logical way of approaching it?

 

[hgfalling]

That should work. What did you get for the derivative after making that change?

 

[cal.queen92]

If I continue, I get:

(1/y)(dy/dx) = ((1/10)+((1/(sinx))(cosx)))

then

(1/y)(dy/dx) = ((1/10)+(cotx))

Then multiplying both sides by y:

dy/dx = (10(sin(x))^1)((1/10)+(cotx))

If I fill in the f(1) then it looks like:

dy/dx = (10(sin(1))^1)((1/10)+(cot(1))

But, another thing that puzzles me is that on a unit circle, Sin(1) = 0 right? So wouldn't the entire answer turn to 0 sine everything is multiplied by Sin(1)?

 

[hgfalling]

\log 10 is a constant, not a function of x. You still need to use product and chain rules when differentiating x \log (\sin x).

And, no, \sin 1 isn't 0. \sin 0, \sin \pi, \sin 2\pi, etc, are zero.

 

[cal.queen92]

Okay, so if ln10 is a constant than its derivative goes to zero. Now I am getting an answer that looks like:

lny = ln10 + xln(sinx)

(1/y)(dy/dx) = 1*ln(sinx) + (1/(sinx))(cosx)(x)

(1/y)(dy/dx) = ln(sinx) + (xcosx/(sinx))

then multiplying both sides by y:

(dy/dx) = (ln(sinx) + (xcosx/(sinx)))(10(sinx)^x)

Then multiplying through:

(10ln(sinx)^(x+1)) + ((10x(cosx)(sinx)^x)/(sinx))

Which is actually very close to the answer:

(10(sinx)^(x+1))(ln(sinx))+(10x(cosx)(sinx)^x))/(sinx)

But these answers aren't exactly the same, I must be missing something.

 

[hgfalling]

then multiplying both sides by y:

(dy/dx) = (ln(sinx) + (xcosx/(sinx)))(10(sinx)^x)

Then multiplying through:

(10ln(sinx)^(x+1)) + ((10x(cosx)(sinx)^x)/(sinx))

ln(sin(x)) and 10(sin x)^x can't be multiplied that way. ln (sin x)*10 (sin x)^x is simplified already.

 

[cal.queen92]

Ohhhh okay, so now the answer looks like:

((ln(sinx))(10(sinx)^(x+1)))+((10x(cosx)(sinx)^x)/(sinx))

But in the actual answer, the whole thing is over sinx, where this answer isn't, it's so close!

 

[hgfalling]

Well you left the x+1 in the exponent. So you'd need a sin x in the denominator to make that right.

 

[cal.queen92]

Okay, so if I did not leave the x+1 in the exponent then the sinx in the denominator wouldn't be necessary? Thanks for all your help!