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

Dec 31, 2015

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

Explanation:

Write as:

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

Now, use the power rule with the chain rule:

$\frac{d}{\mathrm{dx}} \left({u}^{-} 2\right) = - 2 {u}^{-} 3$, and $u = 16 x + 3$.

Thus,

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

$\implies \frac{- 2}{16 x + 3} ^ - 3 \cdot 16$

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