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Composition question

If f(0)=1, f'(0)=2, g(0)=3, g'(0)=4 what is (f^2•g)'(0) 

Comments

Are you sure the question is written correctly?

Faith, it's not clear what f^2 means. It could mean f(x) times f(x), but sometimes f^2 also means f(f(x)), i.e. f composed with itself. Do you know which it is?

sorry for the lack of information, I just copied down what my teacher had on the worksheet. 

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2 Answers

Hi Faith,

As I mentioned in my comment above, the notation "f^2" is ambiguous. I'm going to do the problem two ways, depending on what "f^2" means. But since you said in the title of your post that the problem is a composition problem, I'm assuming the black dot means composition. If it means multiplication, the solutions below aren't right; George's answer is right. In fact, you aren't given enough information to do the problem if you interpret the black dot as composition, so I'm assuming you made a mistake: the problem isn't about composition at all. When dealing with problems like these, you have to make sure you understand exactly what the notation means. Usually composition is denoted by an open dot, not a black dot. The black dot usually stands for multiplication.

Solution One: interpreting "f^2" as f(x) times f(x)

We're trying to find the derivative of the function f^2(g(x)) at x=0. So we just need to differentiate f^2(g(x)) = f(g(x)) x f(g(x)), where I use "x" to mean multiplication. To do that, we can use the product rule: the answer will be

(the derivative of f(g(x))) x f(g(x)) + f(g(x)) x (the derivative of f(g(x)))

which is just

2(the derivative of f(g(x)) x f(g(x)).

(You can also get this answer by applying the chain rule directly to the expression f^2(g(x)).)

But what is the derivative of f(g(x))? To find that, we have to use the chain rule: it's f'(g(x)) x g'(x). So the final answer for the derivative is

2f'(g(x)) x g'(x) x f(g(x))

Now we just have to plug in x=0. We get

2f'(g(0)) x g'(0) x f(g(0)) = 2f'(3) x 4 x f(3)

Since you aren't given f'(3) and f(3), you can't find this.

Solution Two: interpreting "f^2" as f(x) composed with itself, i.e. f(f(x))

We want to differentiate the function f(f(g(x))). To do this, we just use the chain rule twice. The derivative will be

f'(f(g(x)) x (the derivative of f(g(x))) = f'(f(g(x)) x f'(g(x)) x g'(x)

Plugging in x=0, we get

f'(f(g(0))) x f'(g(0)) x g'(0) = f'(f(3)) x f'(3) x 4

You also aren't given enough information to do this.

(uv)' = uv' +vu', and remember Mr. Chain.

(f^2•g)'(0) = 2f(f')g + f^2(g'), all evaluated at 0.

2 f(0)(f'(0))(g(0)) + (f(0))^2(g'(0))  =

2(1)(2)(3) + 1^2(4) =  12 + 4 = 16