Derivative of Composite Function

[Kenji Liew]

Homework Statement

\begin{equation}
f(x)=
\begin{cases}
5x+2 &, x \leq 1 \\
3x^2 &, 1<x<2\\
4-x &, x\geq 2
\end{cases}
\end{equation}

\begin{equation}
g(x)=
\begin{cases}
\frac{1}{5}(2+3 cos x) &, x <0 \\
4-sin x &, x \geq 0
\end{cases}
\end{equation}

Find [itex] h = f \circ g [/itex] and then by using chain rule to find the derivative of [itex] f \circ g [/itex] .

Homework Equations

\begin{equation}
h(x)=f\circ g =
\begin{cases}
4+3 cos x &, x <0 \\
sin x &, x \geq 0
\end{cases}
\end{equation}

The Attempt at a Solution

From [itex] h(x) = f \circ g [/itex] above, I directly differentiate and I get the following
\begin{equation}
h'(x)=
\begin{cases}
-3 sin x &, x < 0 \\
cos x &, x > 0
\end{cases}
\end{equation}

If chain rule are requested in finding the derivative, I know the formula is [itex] h'(x) = f' (g(x)) \cdot g'(x) [/itex]. Any idea to find the derivative using chain rule?

Thank you.

 

[LCKurtz]

Homework Statement

\begin{equation}
f(x)=
\begin{cases}
5x+2 &, x \leq 1 \\
3x^2 &, 1<x<2\\
4-x &, x\geq 2
\end{cases}
\end{equation}

\begin{equation}
g(x)=
\begin{cases}
\frac{1}{5}(2+3 cos x) &, x <0 \\
4-sin x &, x \geq 0
\end{cases}
\end{equation}

Find [itex] h = f \circ g [/itex] and then by using chain rule to find the derivative of [itex] f \circ g [/itex] .

Homework Equations

\begin{equation}
h(x)=f\circ g =
\begin{cases}
4+3 cos x &, x <0 \\
sin x &, x \geq 0
\end{cases}
\end{equation}

The Attempt at a Solution

From [itex] h(x) = f \circ g [/itex] above, I directly differentiate and I get the following
\begin{equation}
h'(x)=
\begin{cases}
-3 sin x &, x < 0 \\
cos x &, x > 0
\end{cases}
\end{equation}

If chain rule are requested in finding the derivative, I know the formula is [itex] h'(x) = f' (g(x)) \cdot g'(x) [/itex]. Any idea to find the derivative using chain rule?

Thank you.

It seems overly pedantic to do it explicitly with the chain rule, but if you must do so, just write h(x) as a two piece formula like this:

If x ≥ 0, f(x) = 4-x, g(x) = 4 - sin(x) and [itex]h=f\circ g[/itex]
If x < 0 f(x) = 5x+2, g(x) = (1/5)(2+3cos(x)) and [itex]h=f\circ g[/itex]
Then do the chain rule on each piece.

 

[Kenji Liew]

It seems overly pedantic to do it explicitly with the chain rule, but if you must do so, just write h(x) as a two piece formula like this:

If x ≥ 0, f(x) = 4-x, g(x) = 4 - sin(x) and [itex]h=f\circ g[/itex]
If x < 0 f(x) = 5x+2, g(x) = (1/5)(2+3cos(x)) and [itex]h=f\circ g[/itex]
Then do the chain rule on each piece.

Thank you for the reply. I got the idea from it. This is because my lecturer want us to compare whether the answer obtained from directly differentiate [itex]h(x)[/itex] is the same
as the answer obtained from the chain rule.