# Differentiation

• Sep 16th 2009, 10:52 PM
Naples
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.
http://i32.tinypic.com/fpb4zo.png
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.
• Sep 16th 2009, 11:50 PM
redsoxfan325
Quote:

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.
http://i32.tinypic.com/fpb4zo.png
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: $\frac{d}{dx}\left[f(h(x))\right] = f'(h(x))\cdot h'(x)$, which in this case is $g(x^2)\cdot 2x$, so the answer is D.

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

For number two: Let $x=CB$ and $y=AB$. We want to solve the equation $dx=\frac{3}{4}dy$ for $x$.

We know that $x^2+y^2=40$ so $y=\sqrt{40-x^2}$ and $dy=\frac{-x}{\sqrt{40-x^2}}\,dx$

We can ignore the minus sign because that just indicates direction and does not affect magnitude. Now we need to solve $dx=\frac{3}{4}\cdot\frac{x}{\sqrt{40-x^2}}\,dx$.

Canceling the $dx$'s and cross-multiplying gives us $4\sqrt{40-x^2}=3x \implies 16(40-x^2)=9x^2 \implies 25x^2=640 \implies x=\sqrt{\frac{124}{5}}=\frac{2\sqrt{155}}{5}$