i know that you can use the point slope formula. but what if all they give for is x=1, for y=e^(-x )+ e^(2x-1). do you find the derv of the function and then plug in 1 to X to find y and then use the point slope formula?
i know that you can use the point slope formula. but what if all they give for is x=1, for y=e^(-x )+ e^(2x-1). do you find the derv of the function and then plug in 1 to X to find y and then use the point slope formula?
if you'll find/get the derivative of f(x)=e^(-x )+ e^(2x-1) and look for f'(1);that is,for example, f'(1)=n, then n is the slope of the tangent line at x=1.
no not really. i dont understand how youre getting (2)e^(2x-1)
It's the chain rule property used so often. The derivative of (2x-1) is just 2, so that is why there is a 2 in front now. The exponent of e remains the same.
It's the chain rule property used so often. The derivative of (2x-1) is just 2, so that is why there is a 2 in front now. The exponent of e remains the same.
what about the first part that says e^-x. isnt that equal to -e^-x? what happens to that?
what about the first part that says e^-x. isnt that equal to -e^-x? what happens to that?
I was only referencing the error you made, which was in the 2nd term. Sorry if that was confusing. You are correct that the derivative of e^(-x) is -e^(-x)