Derivative of Squareroot

Question

What would be the derivate of square roots? For example if i have $2 \sqrt{x}$ or $\sqrt{x}$

Im unsure how to derive these and include them especially in something like implicit

Answer 1

$\sqrt x=x^{1/2}$, so you just use the power rule: the derivative is $\frac12x^{-1/2}$.

Answer 2

Let $f(x) = \sqrt{x} = x^{1/2}$.

$$f'(x) = \frac{1}{2} x ^{-1/2}$$

$$f'(x) = \frac{1}{2x^{1/2}} = \frac{1}{2\sqrt{x}}$$

If you post the specific implicit differentiation problem, it may help. The general guideline of writing the square root as a fractional power and then using the power and chain rule appropriately should be fine however. Also, remember that you can simply pull out a constant when dealing with derivatives - see below.

If $g(x) = 2\sqrt{x} = 2x^{1/2}$. Then,

$$g'(x) = 2\cdot\frac{1}{2}x^{-1/2}$$

$$g'(x) = \frac{1}{x^{1/2}} = \frac{1}{\sqrt{x}}$$

Answer 3

The Power Rule says that $\frac{\mathrm{d}}{\mathrm{d}x}x^\alpha=\alpha x^{\alpha-1}$. Applying this to $\sqrt{x}=x^{\frac12}$ gives $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}\sqrt{x} &=\frac{\mathrm{d}}{\mathrm{d}x}x^{\frac12}\\ &=\frac12x^{-\frac12}\\ &=\frac{1}{2\sqrt{x}}\tag{1} \end{align} $$ However, if you are uncomfortable applying the Power Rule to a fractional power, consider applying implicit differentiation to $$ \begin{align} y&=\sqrt{x}\\ y^2&=x\\ 2y\frac{\mathrm{d}y}{\mathrm{d}x}&=1\\ \frac{\mathrm{d}y}{\mathrm{d}x}&=\frac{1}{2y}\\ &=\frac{1}{2\sqrt{x}}\tag{2} \end{align} $$

Answer 4

Let $f(x) = \sqrt{x}$, then $$f'(x) = \lim_{h \to 0} \dfrac{\sqrt{x+h} - \sqrt{x}}{h} = \lim_{h \to 0} \dfrac{\sqrt{x+h} - \sqrt{x}}{h} \times \dfrac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}} = \lim_{x \to 0} \dfrac{x+h-x}{h (\sqrt{x+h} + \sqrt{x})}\\ = \lim_{h \to 0} \dfrac{h}{h (\sqrt{x+h} + \sqrt{x})} = \lim_{h \to 0} \dfrac1{(\sqrt{x+h} + \sqrt{x})} = \dfrac1{2\sqrt{x}}$$ In general, you can use the fact that if $f(x) = x^{t}$, then $f'(x) = tx^{t-1}$.

Taking $t=1/2$, gives us that $f'(x) = \dfrac12 x^{-1/2}$, which is the same as we obtained above.

Also, recall that $\dfrac{d (c f(x))}{dx} = c \dfrac{df(x)}{dx}$. Hence, you can pull out the constant and then differentiate it.