Some more logarithm stuff giving me trouble

[1MileCrash]

Homework Statement

y = 2^(sin x)

Find derivative.

Homework Equations

The Attempt at a Solution

[itex]y = 2^{sin x}[/itex]
[itex]ln y = ln (2^{sin x})[/itex]
[itex]ln y = sin x ln(2)[/itex]
[itex]y' = [ln(2)] (cos x) (2^{sin x})[/itex]

Yet, according to some rule in the book:

a^x derivative is ln a times a^x, or in this case ln(2)2^(sin x)

But usually these rules don't conflict with other valid ways of coming to the answer. Wolfram shows me my answer, mathway shows me the book's answer.

 

[Pengwuino]

What you have is correct. Remember, you need to use the chain rule.

[tex]{{d}\over{dx}}\left( a^{f(x)} \right) = ln(a)a^{f(x)}f'(x)[/tex]

just as if it were something simpler like [itex]y = e^{x^2}[/itex]

 

[1MileCrash]

Thank you, I was getting frustrated. Another textbook error, I suppose I shouldn't be surprised.

 

[Ray Vickson]

Thank you, I was getting frustrated. Another textbook error, I suppose I shouldn't be surprised.

I think you are mis-interpreting what the book said. According to you, the book said (d/dx) a^x = ln(a) * a^x, and that is perfectly true. However, your question was different. Are you saying that the book's answer for this specific problem, (d/dx) 2^(sin(x)) is wrong? Of course, books DO make mistakes, but I cannot figure out from what you said whether that is the case here, or not.

RGV

 

[HallsofIvy]

Homework Statement

y = 2^(sin x)

Find derivative.

Homework Equations

The Attempt at a Solution

[itex]y = 2^{sin x}[/itex]
[itex]ln y = ln (2^{sin x})[/itex]
[itex]ln y = sin x ln(2)[/itex]
[itex]y' = [ln(2)] (cos x) (2^{sin x})[/itex]

Yet, according to some rule in the book:

a^x derivative is ln a times a^x, or in this case ln(2)2^(sin x)

No, in this case, ln(2)2^(sin x) times the derivative of sin(x) so the derivative is ln(2)2^(sin(x))cos(x) exactly as you have.

But usually these rules don't conflict with other valid ways of coming to the answer. Wolfram shows me my answer, mathway shows me the book's answer.

 

[1MileCrash]

I think you are mis-interpreting what the book said. According to you, the book said (d/dx) a^x = ln(a) * a^x, and that is perfectly true. However, your question was different. Are you saying that the book's answer for this specific problem, (d/dx) 2^(sin(x)) is wrong? Of course, books DO make mistakes, but I cannot figure out from what you said whether that is the case here, or not.

RGV

I'm sorry, I meant both actually.

The book referenced the "box" where that rule was written (right above, on the same page) and then explicitly wrote the answer as n(2)2^(sin x) in accordance with that rule.

I understand the rule is true, but it doesn't nullify the chain rule if I have some function for the exponent. That's what I gathered from the book's example, and that's what was the source of my frustration.

As for mathway...

[PLAIN]http://img263.imageshack.us/img263/1599/mathwaywrong.png

 

[HallsofIvy]

Then mathway is wrong. The derivative of [itex]2^{sin(x)}[/itex] is [itex]2^{sin(x)}ln(2)cos(x)[/itex]. The cos(x), the derivative of sin(x), is missing from what you show.

 

[1MileCrash]

Then mathway is wrong. The derivative of [itex]2^{sin(x)}[/itex] is [itex]2^{sin(x)}ln(2)cos(x)[/itex]. The cos(x), the derivative of sin(x), is missing from what you show.

Yes, this has been established.