If f(x)= csc 7 x  and g(x) = 3x^2 -5 , how do you differentiate f(g(x))  using the chain rule?

Feb 13, 2016

$\frac{d}{\mathrm{dx}} \left(f \left(g \left(x\right)\right)\right) = - 42 x \csc \left(21 {x}^{2} - 35\right) \cot \left(21 {x}^{2} - 35\right)$

Explanation:

First, note the chain rule states that

$\frac{d}{\mathrm{dx}} \left[f \left(g \left(x\right)\right)\right] = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$

Let's focus on the first part, $f ' \left(g \left(x\right)\right)$.

We must find $f ' \left(x\right)$. Ironically, in order to do so, the chain rule must be used once more.

In the case of a cosecant function, the chain rule states that

$\frac{d}{\mathrm{dx}} \left(\csc \left(h \left(x\right)\right)\right) = - \csc \left(h \left(x\right)\right) \cot \left(h \left(x\right)\right) \cdot h ' \left(x\right)$

Thus, since in $\csc \left(7 x\right)$ we see that $h \left(x\right) = 7 x$, and $h ' \left(x\right) = 7$,

$f ' \left(x\right) = - 7 \csc \left(7 x\right) \cot \left(7 x\right)$

Thus to find $f ' \left(g \left(x\right)\right)$ we plug $g \left(x\right)$ into every $x$ in $f ' \left(x\right)$:

$f ' \left(g \left(x\right)\right) = - 7 \csc \left(21 {x}^{2} - 35\right) \cot \left(21 {x}^{2} - 35\right)$

Now, we should find the second term of the original chain rule expression, $g ' \left(x\right)$. This requires only the power rule.

$g ' \left(x\right) = 6 x$

Multiplying $f ' \left(g \left(x\right)\right)$ and $g ' \left(x\right)$ we see that the derivative of the entire composite function is

$\frac{d}{\mathrm{dx}} \left(f \left(g \left(x\right)\right)\right) = - 42 x \csc \left(21 {x}^{2} - 35\right) \cot \left(21 {x}^{2} - 35\right)$