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NCERT Exemplar · Q68

Q.If y=x+1xy = \sqrt{x} + \dfrac{1}{\sqrt{x}}, then dydx\dfrac{dy}{dx} at x=1x = 1 is
(A) 11
(B) 12\dfrac{1}{2}
(C) 12\dfrac{1}{\sqrt{2}}
(D) 00

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The derivative of y=x+1xy = \sqrt{x} + \frac{1}{\sqrt{x}} is found by rewriting each term as a power of xx, differentiating term-by-term, then evaluating at x=1x=1. The result is 00.

The key idea here is that derivative at a point means: first find the general derivative function dydx\frac{dy}{dx} (the rate of change of yy with respect to xx), then plug in the specific xx-value. You cannot substitute x=1x=1 into the original expression and then differentiate — that would give zero for a constant, which misses the point entirely.

Let’s rewrite the function in a form that’s easy to differentiate. Recall that x=x1/2\sqrt{x} = x^{1/2} and 1x=x−1/2\frac{1}{\sqrt{x}} = x^{-1/2}. So:

y=x1/2+x−1/2y = x^{1/2} + x^{-1/2}

Now we differentiate term by term using the power rule: ddxxn=nxn−1\frac{d}{dx} x^n = n x^{n-1}.

  1. Differentiate x1/2x^{1/2}:

ddxx1/2=12x1/2−1=12x−1/2\frac{d}{dx} x^{1/2} = \frac{1}{2} x^{1/2 - 1} = \frac{1}{2} x^{-1/2}

  1. Differentiate x−1/2x^{-1/2}:

ddxx−1/2=−12x−1/2−1=−12x−3/2\frac{d}{dx} x^{-1/2} = -\frac{1}{2} x^{-1/2 - 1} = -\frac{1}{2} x^{-3/2}

So the derivative function is:

dydx=12x−1/2−12x−3/2\frac{dy}{dx} = \frac{1}{2} x^{-1/2} - \frac{1}{2} x^{-3/2}

We can factor 12x−3/2\frac{1}{2} x^{-3/2} out if we like, but it’s not necessary for evaluation.

  1. Now evaluate at x=1x = 1:

x−1/2=1−1/2=1andx−3/2=1−3/2=1x^{-1/2} = 1^{-1/2} = 1 \quad \text{and} \quad x^{-3/2} = 1^{-3/2} = 1

So:

dydx∣x=1=12(1)−12(1)=0\left.\frac{dy}{dx}\right|_{x=1} = \frac{1}{2}(1) - \frac{1}{2}(1) = 0 …

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