L'Hospital's Rule/derivative issue

[tgrrrrr1976]

I'm working on a problem that uses L'Hospital's Rule. I'm supposed to find the limit of (3^(sinx) - 1)/x as x approaches 0. Would not the derivative of 3^(sinx) -1 be 3 raised to the power of sinx, all of which would be multiplied by cosx?
For some reason the book I'm using said that it is 3 raised to the power of sinx multiplied by LN3 multiplied by cosx.

 

[wjm11]

Would not the derivative of 3^(sinx) -1 be 3 raised to the power of sinx, all of which would be multiplied by cosx?
For some reason the book I'm using said that it is 3 raised to the power of sinx multiplied by LN3 multiplied by cosx.

Your book is correct. What you have overlooked is that you are finding the derivative of an exponential function. In your table of commonly used derivatives, look for something like the derivative of "a^u", where "a" is a constant and "u" is some function of x. You then understand where the "ln3" part comes from.

 

[tgrrrrr1976]

I think I see now..the example I was looking at from my notes was e^(x^2 + 5x+2) ..which differentiated to ( e^(x^2 + 5x+2))(x^2 + 5x+2). This is a special case isn't it? It works the way it does because it's already in base e? Since the other one is not, then I need to change it to base e in order to be able to differentiate it? (I'm probably butchering the terms, I apologize)

 

[wjm11]

You don't need to change into base e. It's just that if your base number is e, then lne =1, so that term doesn't appear in your derivative/answer. If the base number is something other than e, (call it "a") you get the ln"a" term in your derivative.

 

[tgrrrrr1976]

OH! I see now. Thanks a lot!