2 the table below gives values for differentiable functions and g use the values to complete par

Step 1: We are given that h(x) = g(x)/f(x). We want to find h'(1). We can use the quotient ru

2 the table below gives values for differentiable functions...

2. The table below gives values for differentiable functions ƒ and g. Use the values to complete parts (a)-(d). X -3 -2 -1 0 1 2 3 f(x) 3 2 1 -1 7 1 2 f'(x) -1 -1 4 2 1 -2 3 g(x) -3 -2 2 1 -1 -2 1 g'(x) 1 -1 -2 -3 -2 -1 2 (a) (2 points) If h(x) = g(x)' f(x), find h'(1). (b) (3 points) If j(x) = x2 f(x)g(x), find j'(-2). (c) (3 points) If k(x) = [f(x)]³ g(x), find k'(0). (d) (3 points) If m(x) = log10(f(x)), find m'(3).

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00:01 In this problem, the functions f and g have continuous derivatives, and the table gives us values of f, f, prime, g, and g prime at selected values of x.

00:16 So for part a, we want to determine h prime of 3 if you know that h of x is equal to g of f of x.

00:35 So in order to find h prime, we're going to have to use our chain rule.

00:42 So h prime of x is equal to g prime of f of x times f prime of x.

00:53 So therefore, h prime of 3 would equal g prime of f of 3 times f prime of 3.

01:04 So the first thing we're going to do is evaluate f of 3.

01:09 So when x is 3, then f of 3 would be 4.

01:14 So we'll have g prime of 4 times f prime of 3.

01:21 And then g prime of 4 says when x is 4, then g prime would be 3.

01:31 And f prime of 3 means when x is 3, f prime of 3, prime is 1.

01:40 So our answer here would be 3.

01:44 I'm just going to clean up our chart here so that we can start our next problem.

01:52 So for part b, we want to calculate m prime of 2 if you know that m of x equals f of x squared.

02:10 So once again, we're applying our chain rule.

02:13 So m prime equals f prime of x squared times 2x.

02:25 So therefore, m prime of 2 would equal f prime of 2 squared times 2 times 2.

02:43 Times 4.

02:45 And f prime of 4 means when x is 4, f prime is negative 6.

02:52 So we'd get negative 6 times 4 or negative 24...

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