# How do you differentiate f(x)=sqrt(x-(3x+5)^2) using the chain rule.?

Jan 22, 2016

$f ' \left(x\right) = - \frac{18 x + 29}{2 \sqrt{- 9 {x}^{2} - 29 x + 25}}$

#### Explanation:

This will be simpler if the terms inside the square root are simplified.

$f \left(x\right) = \sqrt{x - \left(9 {x}^{2} + 30 x + 25\right)} = \sqrt{- 9 {x}^{2} - 29 x + 25}$

Now, to differentiate a square root function, we should treat it as having a fractional exponent.

$f \left(x\right) = {\left(- 9 {x}^{2} - 29 x + 25\right)}^{\frac{1}{2}}$

According to the chain rule, which we will have to use, $\frac{d}{\mathrm{dx}} \left[{u}^{\frac{1}{2}}\right] = \frac{1}{2} {u}^{- \frac{1}{2}} \cdot u '$, and we have $u = - 9 {x}^{2} - 29 x + 25$.

Thus,

$f ' \left(x\right) = \frac{1}{2} {\left(- 9 {x}^{2} - 29 x + 25\right)}^{- \frac{1}{2}} \cdot \frac{d}{\mathrm{dx}} \left[- 9 {x}^{2} - 29 x + 25\right]$

$f ' \left(x\right) = \frac{1}{2 {\left(- 9 {x}^{2} - 29 x + 25\right)}^{\frac{1}{2}}} \cdot \left(- 18 x - 29\right)$

$f ' \left(x\right) = - \frac{18 x + 29}{2 \sqrt{- 9 {x}^{2} - 29 x + 25}}$