Finding the derivative of m(x)=-e^xcosx

[Buzzlastyear]

Homework Statement

Find the derivative of m(x)=-e^x*cosx at x=1 graphically and algebraically.

Homework Equations

The Attempt at a Solution

So first i just attempted to find the derivative:

m'(x)=-xe^(x-1)*(-sinx)----> is this correct?

so to find at x=1 i just sub in 1.

m'(1)=-(1)e^(1-1)*(-sin1)
m'(1)=-1e^0*(-sin1)
m'(1)=0.01745

Is this correct? When i tried graphing the derivative i got a different answer so it made me second guess my original answer.

Any help is very much appreciated, thank you.

 

[HallsofIvy]

Homework Statement

Find the derivative of m(x)=-e^x*cosx at x=1 graphically and algebraically.

Homework Equations

The Attempt at a Solution

So first i just attempted to find the derivative:

m'(x)=-xe^(x-1)*(-sinx)----> is this correct?

No, it's not. You are mis-using some basic properties of the derivative. In the first place the "product rule": the derivative of fg is (fg)'= f'g+ fg', not just " f'g' " as you have here.

Second, the rule " (xⁿ)'= n x^n-1" applies only when the variable is the base and the exponent is a constant. With e^x or, more generally, a^x, for any positive a, that rule does not apply. Look up the derivative of e^x sdecifically.

so to find at x=1 i just sub in 1.

m'(1)=-(1)e^(1-1)*(-sin1)
m'(1)=-1e^0*(-sin1)
m'(1)=0.01745

Is this correct? When i tried graphing the derivative i got a different answer so it made me second guess my original answer.

Any help is very much appreciated, thank you.

 

[Buzzlastyear]

Okay, thank you. I had not learned about the derivative of exponential functions but looked it up. So is -e^x still just -e^x?

 

[Buzzlastyear]

And as for using the product rule, would this be correct?

m'(x)=-e^x*(cosx) + (-sinx)(-e^x)

assuming my previous post is correct?

 

[HallsofIvy]

Yes, f(x)= e^x has the very nice property that its derivative is just f'(x)= e^x again. That's why the number "e" has a special symbol.

 

[Buzzlastyear]

Okay, thanks very much for your help.