Differentiation of a Square root

Question

Can the following be differentiated using first principles or the power rule?

$$y = (x^3+6x^2+3x-10)^{0.5}$$

I know I have to get it into an expression where each term on the right hand side has its own power, instead of the whole thing having a power; but get stuck there.

Any help will be appreciated.

$d/dx \left(f(x)^{1/2}\right) = 1/2 ~ \left(f(x)\right)^{-1/2}~f'(x)$ — Moo, Feb 21 '16 at 05:28
Logarithmic differentiation is also useful. $\ln y = \dfrac 12 \ln(x^3+6x^2+3x-10)$, $\dfrac {y'}y =\dfrac 12 \dfrac{3x^2 + 12x + 3}{x^3+6x^2+3x-10}$, $y' = \dfrac y2 \dfrac{3x^2 + 12x + 3}{x^3+6x^2+3x-10} = \dfrac{3 x^2+12 x+3}{2 \sqrt{x^3+6 x^2+3 x-10}}$ — Steven Alexis Gregory, Feb 21 '16 at 06:19

score 3 · Answer 1 · answered Feb 21 '16 at 05:30

3

Yes you use power rule and chain rule.

$ y' = \frac{1}{2}(x^3+6x^2+3x-10)^{-\frac{1}{2}} \frac{d}{dx}\left((x^3+6x^2+3x-10)\right) = \dfrac{3x^2+12x+3}{2(x^3+6x^2+3x-10)^\frac{1}{2}} $

answered Feb 21 '16 at 05:30

alpastor

1,460

Jaume Oliver Lafont · Accepted Answer · 2016-02-21T06:23:09.100

3

You may take the root out of the RHS by squaring both sides of the equation.

$$ \begin{align} y &= (x^3+6x^2+3x-10)^\frac{1}{2} \\ y^2 &=x^3+6x^2+3x-10 \end{align}$$

Now differentiate to obtain $$\begin{align} \frac{d}{dx}(y^2) &=\frac{d}{dx}(x^3+6x^2+3x-10) \\ 2y\frac{dy}{dx} &= (3x^2+12x+3) \\ y'=\frac{dy}{dx} &=\frac{3x^2+12x+3}{2y}=\frac{3x^2+12x+3}{2\left(x^3+6x^2+3x-10\right)^\frac{1}{2}} \end{align}$$

edited Feb 21 '16 at 06:23

answered Feb 21 '16 at 06:08

Jaume Oliver Lafont

5,212

1

@GoodChessPlayer: Note that this method (implicit differentiation) is in fact one way we can easily prove that $\frac{d}{dx}( \sqrt{x} ) = \frac{1}{2\sqrt{x}}$. With not too much difficulty you can also get $\frac{d}{dx}( x^r ) = r x^{r-1}$ for any $x > 0$ and rational $r$. Beyond this you will need some more advanced real analysis to obtain the same rule for all real $r$. Anyway it is of course good to remember the result so that you don't have to rederive it every time. – user21820 Feb 21 '16 at 09:22

score 1 · Answer 3 · answered Feb 21 '16 at 05:29

1

it is by the power rule $$\frac{1}{2}(x^3+6x^2+3x-1)^{-1/2}(3x^2+12x+3)$$ and the chain rule

answered Feb 21 '16 at 05:29

Dr. Sonnhard Graubner

95,283

score 1 · Answer 4 · edited Oct 15 '23 at 07:53

1

$$\begin{align*} y &=\sqrt{x^3+6x^2+3x-10} \\ \quad \quad \frac{\mathrm{d}y}{\mathrm{d}x} &=\frac{\mathrm{d}{(\sqrt{x^3+6x^2+3x-10})}}{\mathrm{d}x} \\ &=\frac{\mathrm{d}{(\sqrt{x^3+6x^2+3x-10})}}{\mathrm{d}{(x^3+6x^2+3x-10)}} \cdot \frac{\mathrm{d}{(x^3+6x^2+3x-10)}}{\mathrm{d}x} \\ &=\frac{1}{2\sqrt{x^3+6x^2+3x-10}} \cdot (3x^2+12x+3) \\ &=\frac{3x^2+12x+3}{2\sqrt{x^3+6x^2+3x-10}} \end{align*}$$

edited Oct 15 '23 at 07:53

IraeVid

3,216

answered Feb 21 '16 at 06:41

MathIsNice1729

4,510

Using which rules do you calculate this? And do those rules explain the factor two that suddenly appears? – hkBst Feb 21 '16 at 09:25
@hkBst I used the chain rule and the rule of polynomial differentiation (i.e. the power rule). – MathIsNice1729 Feb 21 '16 at 10:05

Differentiation of a Square root

4 Answers4