Question

y=(lx+m)^1/2 , then find dy/dx

Answer 1

Answer :

Given that,

y = $\sqrt{lx+m}$

Now, differentiating both sides with respect to x, we get

$\frac{dy}{dx}= \frac{d}{dx} (lx + m)^{\frac{1}{2}}$

= $\frac{1}{2} (lx + m)^{\frac{1}{2} - 1} \times \frac{d}{dx} (lx + m)$

= $\frac{1}{2} (lx + m)^{-\frac{1}{2}} \times l$

= $\frac{l}{2} \frac{1}{(lx+m)^{\frac{1}{2}}}$

= $\frac{l}{2} \frac{1}{y}$

= $\frac{l}{2y}$

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