# How do you differentiate f(x)=1/(16x+3)^2 using the quotient rule?

Jul 27, 2017

$f ' \left(x\right) = - \frac{32}{16 x + 3} ^ 3$

#### Explanation:

$\text{given " f(x)=(g(x))/(h(x))" then}$

$f ' \left(x\right) = \frac{h \left(x\right) g ' \left(x\right) - g \left(x\right) h ' \left(x\right)}{h \left(x\right)} ^ 2 \leftarrow \text{ quotient rule}$

$g \left(x\right) = 1 \Rightarrow g ' \left(x\right) = 0$

$h \left(x\right) = {\left(16 x + 3\right)}^{2}$

$\Rightarrow h ' \left(x\right) = 2 \left(16 x + 3\right) .16 = 32 \left(16 x + 3\right) \leftarrow \text{ chain rule}$

$\Rightarrow f ' \left(x\right) = \frac{{\left(16 x + 3\right)}^{2} .0 - 32 \left(16 x + 3\right)}{16 x + 3} ^ 4$

$\textcolor{w h i t e}{\Rightarrow f ' \left(x\right)} = - \frac{32}{16 x + 3} ^ 3$

Jul 27, 2017

$- \frac{32}{16 x + 3} ^ 3$

#### Explanation:

We first find the derivative using the quotient rule:

$\frac{f ' \left(g\right) - g ' \left(f\right)}{{g}^{2}}$

$f \left(x\right) = \frac{1}{16 x + 3} ^ 2$

Where $f = 1$ and $g = {\left(16 x + 3\right)}^{2}$

But we have $16 x + {3}^{2}$ we proceed to differentiate it using the chain rule, where we take the derivative of the "outside" and multiply it by the derivative of the "inside". Knowing this we can proceed now find the derivative:

$\frac{d}{\mathrm{dx}} = \frac{0 {\left(16 x + 3\right)}^{2} - 2 \left(16 x + 3\right) \left(16\right) \left(1\right)}{{\left(16 x + 3\right)}^{2}} ^ 2$

Now we clean it up:

$\frac{d}{\mathrm{dx}} = \frac{\cancel{0 {\left(16 x + 3\right)}^{2}} - 32 \left(16 x + 3\right)}{16 x + 3} ^ 4$

We can cancel out just one $16 x + 3$

We now have:

$\frac{d}{\mathrm{dx}} = - \frac{32 \left(\cancel{16 x + 3}\right)}{16 x + 3} ^ \left(4 - 1\right)$

$\frac{d}{\mathrm{dx}} = - \frac{32}{16 x + 3} ^ 3$