# Can I get help with this graph? It involves derivatives and differentiation. I got h'(2) right but need help with h'(1) and h'(3)

Jun 13, 2018

Given that $h \left(x\right) = f \left(x\right) \cdot g \left(x\right)$, we know from the product rule that $h ' \left(x\right) = f ' \left(x\right) g \left(x\right) + f \left(x\right) g ' \left(x\right)$.

Therefore

$h ' \left(1\right) = f ' \left(1\right) g \left(1\right) + f \left(1\right) g ' \left(1\right)$

The values of $f ' \left(1\right)$ and $g ' \left(1\right)$ will be determined by taking the slope of the graphs at those points.

$f ' \left(1\right) = \frac{3 - 0}{2 - 0} = \frac{3}{2}$
$g ' \left(1\right) = \frac{1 - 0}{0 - 4} = - \frac{1}{4}$

To determine the values of $f \left(1\right) \mathmr{and} g \left(1\right)$, knowing the equation of the lines would be helpful.

For f(x): $y - 0 = \frac{3}{2} \left(x - 0\right) \to y = \frac{3}{2} x$
For g(x): $y - 1 = - \frac{1}{4} \left(x - 0\right) \to y = - \frac{1}{4} x + 1$

Therefore

$f \left(1\right) = \frac{3}{2} \left(1\right) = \frac{3}{2}$
$g \left(1\right) = - \frac{1}{4} \left(1\right) + 1 = \frac{3}{4}$

Piecing all this together:

$h ' \left(1\right) = \frac{3}{2} \left(\frac{3}{4}\right) + \left(- \frac{1}{4}\right) \left(\frac{3}{2}\right) = \frac{9}{8} - \frac{3}{8} = \frac{6}{8} = \frac{3}{4}$

The same process is required for $h \left(3\right)$. I'll leave that up to you.

Hopefully this helps!