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Calculus/Calculus- Derivatives

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Expert: Scotto - 8/27/2008


Question
I was wondering if you could explain what the dx/dy is and how to use it.

And how would you find the derivative of say:
y= (2x-7)^3

And how would you evaluate the derivative fo the function at an indicated point like:
y= 37-sec^3(2x)        and the points are (0,36)
The 2x isn't with the cubed...

That would be so helpful! Thankies! :}

Answer
The derivative of (f(x))^n is n(f(x))^(n-1)f'(x).  Here, f(x)=2x-7 and n is 3.

At (0,36), we have 37-sec^3(0)=36, so all that says is what's true since the sec(0) is 1 and therefore the cube is 1, so 37-1=36 is true as well.

On the function sec^3(2x), 2x is computed, then the secant is taken, and then the value gotten from the secant is cubed.  It looked like maybe you were a little unclear when you said the 2x wasn't cubed.

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Scotto

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Any kind of calculus question you want. I also have answered some questions in Physics (mass, momentum, falling bodies), Chemistry (charge, reactions, symbols, molecules), and Biology.

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Experience in the area: I have tutored students in all areas of mathematics for over 25 years. Education/Credentials: BSand MS in Mathematics from Oregon State University, where I completed sophomore course in Physics and Chemistry. I received both degrees with high honors. Awards and Honors: I have passed Actuarial tests 100, 110, and 135.

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