Assignment 4

6 February 2018 in MAE 113: Calculus for the Liberal Arts | 2 comments

Assignment 4 was handed out in class. You can see a copy here with the additional solutions here.

I had a question about some of the solutions.

In Question 4(g), why does \pi - 1 as an exponent necessitate using x as a denominator.

It is because x^{\pi - 1} = x^\pi \cdot x^{-1} = \dfrac{x^\pi}{x}

    \begin{align*} f(x) &= 3x^\pi - \sqrt{2}x^{\sqrt{2}}\\ f'(x) &= 3(\pi x^{\pi - 1}) - \sqrt{2}(\sqrt{2} x^{\sqrt{2} - 1})\\ &= 3\pi x^{\pi - 1} - 2x^{\sqrt{2} - 1}\\ &= 3\pi x^{\pi} x^{-1} - 2x \sqrt{2} x^{-1}\\ &= \dfrac{3\pi x^{\pi}}{x} - \dfrac{2 x^{\sqrt{2}}}{x}\\ &= \dfrac{3\pi x^{\pi} - 2x^{\sqrt{2}}}{x} \end{align*}

Tags: , ,

  1. Lisa’s avatar

    Lisa on 12 February 2018 at 9:35 pm

    Thanks for this, makes sense now. Would this be for any variable? Ex. y-1 as exponent?

    Reply

  2. Richard Kohar’s avatar

    Richard Kohar on 13 February 2018 at 11:50 am

    Yes. For example, x^{y - 1} = x^y \cdot x^{-1} = \dfrac{x^y}{x}.

    Reply

Reply

Your email address will not be published. Required fields are marked *