# Derivative HW

1. Mar 19, 2004

### physicszman

Hi, these are just a few problems from my assignment I am having a hard time with. Any help is appreciated.

1) consider function f(x) = 1/3X^3 - 1/2^2 - 6x + 4

a) find the points on the graph at which the tangent lines are horizontal.
b) find the slope of the line tangent to the curve at x = 3
f) write an equation of this tangent line

2) After t years the population P is given by P(t) = 2000t^2 + 10000t

what is the growth rate at t = 10?

a) f(x) = 1/2x^1/2 + 2/3x^1/4 - x^1/5
f'(x) = 1/4x^-1/2 + 1/6x^-3/4 - 1/5x^-4/5

b) y = 5x^1/3
y'= 5/3x^-2/3

c) g(x) = x^1/3 - x^1/3
g'(x) = 1/3x^-2/3 - 1/3x^-2/3

d) y = x^-3/4 - 3x^2/3
y'= -3/4x^-7/4 - 2x^-1/3

e) h(x) = x/2 + 2/x
h'(x) = ?

f) y = (3x^2 - 5x +7)/4
y'= ?

g)f(x) = -1/3x^3 + 4x^2 - 5x + 6
f'(x)= -x^2 + 8x - 5

h) g(x) = (x^3 + x^5)/(x^2)
g'(x) = ?

Thanks again!!

2. Mar 20, 2004

### Chen

1) Did you try finding the derivative of this (rather simple) function?
a) A horizontal line is of the form [y = k], i.e its slope is zero. And since the derivative of a function gives you the slope of the tangent to the function, when the derivative is zero the tangent line is horizontal.
b) To find the slope of the tangent line at x = 3, simply find the function's derivative and substitute x for 3. The number you get is the tangent line's slope at that point.
f) What do you need to write the equation of a line? In this case, its slope (m), and a point on it ($$x_1, y_1$$). You already have the slope, you found it in b). The only thing you are missing is a point on the line, and that you can find easily because you know the line is tangent to the function at x = 3. So find f(x = 3) and you will have the ($$x_1, y_1$$) coordinates of a point on the line. And then it's just $$y = m(x - x_1) + y_1$$.

2) What is the growth rate? It is the difference in the population in a given time period. What is the slope of a line? It is the difference in Y in a given delta X. But since P(t) is a parabola, the slope changes and is not constant. So to find the slope, i.e the growth rate, after 10 years, find the derivative of P(t) and substitute t for 10.

3. Mar 20, 2004

### Chen

a), b), c), d) and g) all look correct.

e) $$\frac{2}{x}$$ can also be written as $$2x^{-1}$$. And you already know how to find the derivative that.

f) The derivative of $$\frac{f(x)}{k}$$, where k is a constant that doesn't depend on x, is the same as the derivative of $$\frac{1}{k}f(x)$$ which is $$\frac{1}{k}f'(x)$$.

h) The derivative of $$\frac{f(x)}{g(x)}$$ is $$\frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}$$. In words, it is the derivative of f(x) times g(x) (itself, not its derivative) minus the derivative of g(x) times f(x) divided by g(x) to the power of two. Try solving it, the answer should be:
$$g'(x) = 3x^2 + 1$$