Calculus Problems!?

Question

I have some questions about some calculus problems. If anyone could explain to me how they are done it would be much appreciated:
A) If f(x) = x^3, find an expression for d/dx [f(g(x))].

B) Find dy/dx for sin(xy) = y.

C) Consider the curve given by y^3 + 5x^2y - 18 = 0.
Write an equation for the line tangent to the curve at the point (1,2).

D) If x and y are both differentiable functions of t, and xy = 20, find
x'(t) when y'(t) = 10 and x = 2.

E) Given that t, k, and a are constants, and that f(x) = a - 2kx, find f'(t).

F) If f'(a) does NOT exist, which of the following MUST be true? If false, show an example of a graph where it is false.
1. f(x)is discontinuousat x = a
2. the limit as x-->a of f(x) does not exist
3. f(x) has a vertical tangent at x = a
4. f(x) has a "hole" at x = a
5. none of the above is necessarily true.

Thanks for any help you can give me!!!

Answer 1

A: This is a simple application of the chain rule:

d(f(g(x)))/dx = f'(g(x)) g'(x) = 3g(x)²g'(x)

B: Here we differentiate implicitly, and then solve for dy/dx:

sin (xy) = y
d(sin (xy))/dx = dy/dx
cos (xy) d(xy)/dx = dy/dx
cos (xy) (y + x dy/dx) = dy/dx
y cos (xy) + x cos (xy) dy/dx = dy/dx
y cos (xy) = dy/dx - x cos (xy) dy/dx
y cos (xy) = (1 - x cos (xy)) dy/dx
dy/dx = y cos (xy)/(1 - x cos (xy))

C: First, we use implicit differentiation to find the slope, then we write the line in point-slope form:

3y² dy/dx + 10xy + 5x² dy/dx = 0
(3y²+5x²) dy/dx = -10xy
dy/dx = -10xy/(3y²+5x²)
Substituting (1, 2):
dy/dx = -20/(12+5) = -20/17

The tangent line is therefore:

y-2 = -20/17 (x-1)
y-2 = -20x/17 + 20/17
y = -20x/17 + 54/17

D: Again, using implicit differentiation:

xy=20
d(xy)/dt = d(20)/dt
y dx/dt + x dy/dt = 0
dx/dt = x/y dy/dt

From our original equation:
xy=20
y=20/x
So:
dx/dt = x²/20 dy/dt
Substituting:
dx/dt = 4/20 * 10 = 2

E: f'(t) = -2k. This is about the simplest derivative in the world

F: None of the above is necessarily true. In fact, a single counterexample refutes all four: the function f(x) = |x| This function:

Has a limit at 0 ([x→0+]lim f(x) = [x→0+]lim f(x) = f(0) = 0)
Is continuous at 0 (per the above)
Does not have a hole at 0 (since f(0) is defined -- specifically, f(0)=0).
Has NO tangent at 0
And is not differentiable at 0 (since [h→0]lim (f(0+h) - f(0))/h = [h→0]lim |h|/h, but [h→0+]lim |h|/h = 1 and [h→0-]lim |h|/h = -1, and 1≠-1).

Answer 2

A) Use the chain rule. Recall that the chain rule states:

[f(g(x))]' = f'(g(x)*g'(x)

B) Implicit differentiation. See C for example.

C) Here you should take an implicit derivative to find dy/dx. In this case, differentiating implicity with respect to x gives:

3y^2 dy/dx + 10xy + 5x^2 dy/dx = 0

Solve this for dy/dx, then plug in (1,2) to get the slope of the tangent line.

D) Take the derivative of both sides of the equation. Plug in appropriate values.

E) Take the derivative, evaluate at t.

F) Consider the absolute value function at a = 0.