Question

𝑓(𝑥) = 𝑥^2 + 1, 𝑔(𝑥) = √𝑥.
Determine the derivative of 𝑖) 𝑓(𝑔(𝑥)), 𝑖𝑖) 𝑔(𝑓(𝑥)),
𝑖𝑖𝑖) 𝑓(𝑓(𝑥)), and 𝑖𝑣) 𝑔(𝑔(𝑥)) with respect to 𝑥.

Answer 1

@Vocaloid When you get a chance

Answer 2

Well
i) derivative of f(g(x)) is
\[((\sqrt{x})^{2}+1)' = (x+1)'\]

Answer 3

Can you find the derivative of x+1?

Answer 4

Suspect they are looking for you to use the Chain Rule and understand composite functions, not just brute force by evaluating the composition first.....though that is a perfectly OK way to do it, tbc.

For the first one, and using the simplest (prime) notation I can think of, take the composition:


\(F(x) = f(g(x))\) <<-- x is the independent single variable

Then, by the Chain Rule:

\(F'(x)= f'(g(x)) \cdot g'(x) \qquad \square\)

For \(f(x) = x^2 + 1\) and \(g(x) = x^{\frac{1}{2}}\):

: \(f'(x) = 2x\),

: \(g'(x) = \frac{1}{2 }x^{-\frac{1}{2}}\)


So:

\( f'(g(x)) = 2(g(x)) = 2 x^{\frac{1}{2}}\)


Thus:

\( F'(x) = f'(g(x)) \cdot g'(x)= 2 x^{\frac{1}{2}} \cdot \frac{1}{2 }x^{-\frac{1}{2}} = ???\)

JUMPING to the **last one**: \(g(g(x)) = \sqrt{\sqrt{x}} = x^{1/4}\), so the derivative is \(\frac{1}{4}x^{- \frac{3}{4}}\) and brute-force looks a simpler way to do this :(

Using the Chain Rule, our composite function is:


\(F(x) = g(g(x))\)

Follows from \(\square\) that:

\(F'(x) = g'(g(x)) \cdot g'(x)\qquad \triangle\)

We already know that \(g'(x) = \frac{1}{2 }x^{-\frac{1}{2}}\)

So:

\(g'(g(x)) = \frac{1}{2 }(g(x))^{-\frac{1}{2}} = \frac{1}{2 }(x^{\frac{1}{2}})^{-\frac{1}{2}} = \frac{1}{2 }x^{-\frac{1}{4}}\)

From \(\triangle\), the Chain Rule says that:

\(F'(x) = \frac{1}{2} x^{-\frac{1}{4}} \cdot \frac{1}{2 }x^{-\frac{1}{2}} = \frac{1}{4} x^{-\frac{3}{4}} \)

So why bother with the chain rule when brute-force is so simpler? Well, try differentiating something like:

\(F(x) = e^{\sin (x^2 + 4)} \qquad F'(x) = ???\)

Break it down with the Chain Rule and it is much simpler:

: \(F(x) = e^{a(x)} \qquad a(x) = \sin (x^2 + 4)\)

: \(a(x) = \sin (b(x)) \qquad b(x) = x^2 + 4\)

That is a straightforward composition that you can differentiate using the Chain Rule

I am guessing that you're supposed to be spreading your wings, math-wise. Maybe the spirit of the question was looking for a different approach.