Question

Differentiate the following functions with respect to x (without using first principle) :

i) (Sinx)^1/2​

Answer 1

We apply chain rule.

$\dfrac {d}{dx}[f(g(x))]=\dfrac {d}{d[g(x)]}[f(g(x))]\cdot\dfrac {d}{dx}[g(x)]$

Let,

$t=\sin x\\\\\\\dfrac {dt}{dx}=\cos x$

Then,

$\dfrac {d}{dx}\left[t^{\frac {1}{2}}\right]=\dfrac {d}{dt}\left[t^{\frac {1}{2}}\right]\cdot\dfrac {dt}{dx}\\\\\\\dfrac {d}{dx}\left[t^{\frac {1}{2}}\right]=\dfrac {1}{2}t^{-\frac {1}{2}}\cdot\dfrac {dt}{dx}\\\\\\\dfrac {d}{dx}\left[t^{\frac {1}{2}}\right]=\dfrac {1}{2t^{\frac {1}{2}}}\cdot\dfrac {dt}{dx}$

That is,

$\underline{\underline {\dfrac {d}{dx}(\sin x)^{\frac {1}{2}}=\dfrac {\cos x}{2(\sin x)^{\frac {1}{2}}}}}$

Answer 2

Answer :

y' = cos x/2√sin x

Step-by-step explanation:

Let y = (sin x)½

To finD the derivative of y w.r.t x

We use the power rule first, followed by the chain rule

$\longrightarrow \sf \: y' = \dfrac{1}{2} \times {sin}^{ \frac{1}{2} - 1 } x \\ \\ \longrightarrow \: \sf \: y' = \dfrac{1}{2 \: \sqrt{sin \: x} } \times \frac{d(sin \: x)}{dx} \\ \\ \longrightarrow \: \sf \: y' = \dfrac{cos \: x}{2 \sqrt{sin \: x} }$