$\displaystyle y = 2 log_7 (5x) + e^{-x^2}$
You can approach this in two ways. First:
$\displaystyle y = log_b(x)$ => $\displaystyle y' = \frac{1}{x ln(b)}$
or simply use the change of base formula:
$\displaystyle log_b(x) = \frac{log_a(x)}{log_a(b)}$
I'm going to use the change of base formula and change the base to e:
$\displaystyle y = 2 log_7 (5x) + e^{-x^2} = 2 \cdot \frac{ln(5x)}{ln(7)} + e^{-x^2}$
Now it's just a matter of using the chain rule:
$\displaystyle y' = \frac{2}{ln(7)} \cdot \frac{1}{5x} \cdot 5 + e^{-x^2} \cdot (-2x)$
$\displaystyle y' = \frac{2}{xln(7)} - 2xe^{-x^2}$
-Dan
$\displaystyle y = (2x^3 + 5)(3x^2 - x)$
There are two ways to do this.
First, multiply out the expression:
$\displaystyle y = 6x^5 - 2x^4 + 15x^2 - 5x$
Then
$\displaystyle y' = 30x^4 - 8x^3 + 30x - 5$
Or use the product rule:
$\displaystyle y = (2x^3 + 5)(3x^2 - x)$
$\displaystyle y' = (6x^2)(3x^2 - x) + (2x^3 + 5)(6x - 1)$
which reduces to the same expression as above.
-Dan
$\displaystyle y = \frac{x^2 - 7}{2x+1} \cdot e^{x^2}$
This is a combination of the quotient and product rules:
$\displaystyle y' = \frac{(2x)(2x + 1) - (x^2 - 7)(2)}{(2x + 1)^2} \cdot e^{x^2} + \frac{x^2 - 7}{2x + 1} \cdot e^{x^2} \cdot (2x)$
$\displaystyle y' = \frac{2x^2 + 2x + 14}{(2x + 1)^2} \cdot e^{x^2} + \frac{2x(x^2 - 7)}{2x + 1} \cdot e^{x^2}$
I'll let you do the rest of the simplifying.
-Dan