Question

# Let P(x)=F(x)G(x) and Q(x)=\frac{F(x)}{G(x)}, where F and G are the functions whose graphs are shown. a) Find P'(2) b) Find Q'(7)

Let $$P(x)=F(x)G(x)$$ and $$Q(x)=\frac{F(x)}{G(x)}$$, where F and G are the functions whose graphs are shown.
a) Find $$P'(2)$$
b) Find $$Q'(7)$$

2021-06-04
Step 1
a) $$P'(2)$$ will be the derivative of P(x) evaluated at $$x=2$$. So first, we take a derivative.
In this section, you should have learned product rule, so $$P'(x)$$ will look like this:
$$P'(x)=F(x)G'(x)+F'(x)G(x)$$
Now we let $$x=2$$
$$P'(2)=F(2)G'(2)+F'(2)G(2)$$
Read these values off the graph.
We see that $$F(2)=3$$ (When its x-value is 2, the curve F has a y-value of 3)
We see that $$G(2)=2$$.
$$F'(2)$$ will be the slope of the curve F at 2. A tangent line at $$F(2)$$ would be horizontal, so $$F'(2)=0$$.
$$G(2)$$ has a slope of $$\frac{1}{2}$$, so $$G'(2)=\frac{1}{2}$$
Now we substitute these values into $$P'(2)$$ to solve:
$$P'(2)=(3)\left(\frac{1}{2}\right)+(0)(2)=\left(\frac{3}{2}\right)+(0)=\frac{3}{2}$$
Step 2
We're playing by the same rules as above-differentiate the equation, read the values off the graph, substitute the values into the function, and solve.
Differentiate (Quotient Rule):
$$Q'(x)=\frac{G(x)F'(x)-F(x)G'(x)}{G(x)^{2}}$$
Read the values off the graph:
$$G(7)=1$$
$$G'(7)=-\frac{2}{3}$$
$$F(7)=5$$
$$F'(7)=\frac{1}{4}$$
Substitute:
$$Q'(7)=\frac{[1\cdot\left(\frac{1}{4}\right)-5\cdot\left(-\frac{2}{3}\right)]}{[1^{2}]}$$
$$Q'(7)=\left[\left(\frac{1}{4}\right)+\left(\frac{10}{3}\right)\right]$$
$$Q'(7)=\left(\frac{43}{12}\right)=3.5833$$