Thread: Differentiation

1. If d/dx(f(x)) = g(x) and if h(x) = x^2, then d/dx(f(h(x)) = ?
A) g(x^2)
B) 2xg(x)
C) g'(x)
D) 2xg(x^2)
E) x^2g(x^2)

I got A,
d/dx(f(h(x))
= g(h(x))
= g(x^2)

Not 100% sure it that's right so I'd like to get someone to check it please.


2.

In the figure, AB is a 40 ft ladder with end A against a vertical wall and end B on level ground. If the ladder is sliding down the wall, what is the distance BC at the instant when B is moving along the ground 3/4 as fast as A is moving down the wall?

I derived pythagoreans theorem to get
2x(dx/dt) + 2y (dy/dt) = 0
The problem is I don't have values for either x or y and I'm not sure how to get them.

Originally Posted by Naples

1. If d/dx(f(x)) = g(x) and if h(x) = x^2, then d/dx(f(h(x)) = ?
A) g(x^2)
B) 2xg(x)
C) g'(x)
D) 2xg(x^2)
E) x^2g(x^2)

I got A,
d/dx(f(h(x))
= g(h(x))
= g(x^2)

Not 100% sure it that's right so I'd like to get someone to check it please.


2.

In the figure, AB is a 40 ft ladder with end A against a vertical wall and end B on level ground. If the ladder is sliding down the wall, what is the distance BC at the instant when B is moving along the ground 3/4 as fast as A is moving down the wall?

I derived pythagoreans theorem to get
2x(dx/dt) + 2y (dy/dt) = 0
The problem is I don't have values for either x or y and I'm not sure how to get them.

Number one is the chain rule: , which in this case is , so the answer is D.

---------------

For number two: Let and . We want to solve the equation for .

We know that so and

We can ignore the minus sign because that just indicates direction and does not affect magnitude. Now we need to solve .

Canceling the 's and cross-multiplying gives us