Good Morning. I've got this really tough multi-step problem that I just can't get. I've gotten the answer key, but just can't get the steps right. Can you help?
My question is:
Suppose that f(x) and g(x) are differentiable everywhere and that f(4)=6, f'(4)=3, g(4)=-2 and g'(4)=10
Part 1: if y=2f(x)g(x), what is y'(4)
Part 2: if y=g(x)/f(x), what is y'(4)
Part 3: Find the equation of the tangent line to 3(f(x))^2 at x=4
Thank you for your assistance,
Using the standard rules for differentiation, these problems really aren't that tough (except maybe 3). The idea is to perform the differentiation to get an expression involving the functions and their derivatives and then to plug in their values at the given value of x.
(N.B. check my algebra and arithmetic!)
1. Using the Product Rule: y' = dy/dx = d[2f(x)･g(x)]/dx = 2[f'･g + f･g']
plugging in the x values --> y' = 2[f'(4)･g(4) + f(4)･g'(4)] = 2[(3)(-2) + (10)(6)] = 108
2. This one uses the Quotient Rule: d[f(x)/g(x)]/dx = (f'･g - f･f')/g^2 --> [(2)(3) - (6)(10)]/(10)^2 = 0.66.
3. This one is asking for the equation of a line, which can be written in the simple form y = m･x + b, where m is the slope of the line and b is the intercept (you should be familiar with this expression). The problem boils down to finding m and b. The slope of the line is just the derivative of y at the given value of x, so
y'(x=4) = 3･d[f(x)^2]/dx = 3･2･f(4) = 6･6 =36 = slope of desired line.
Now we need to find the intercept b. It is defined as the value of the line when it crosses the y axis, which happens when x = 0, i.e., we want to find y(0). To find this, we need to write the equation of the line a little more elaborately so that it includes the the point it shares with the function y = 3f(x)^2. In terms of the slope,
(y - y0)/(x - x0) = m
where x0 = 4 and y0 = 3f(x=4)^2 = 3･(6^2) = 3･36 = 108.
Rearranging, we get
y(x) = m･(x-x0) + y0
so that for x = 0 we have
b = y(0) = -m･x0 + y0 = -36･4 + 108 = -36.
Piece of cake.