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Chapter 6: Problem 34

If \(y=\sqrt{x}+\frac{1}{\sqrt{x}}\), prove that \(2 x \frac{d y}{d x}=\sqrt{x}-\frac{1}{\sqrt{x}}\).

Short Answer

Expert verified
The proof that \(2x \cdot \frac{d y}{d x} = \sqrt{x} - \frac{1}{\sqrt{x}}\) is thus completed.

Step by step solution

01

Differentiate the given function

The given function \(y = \sqrt{x} + \frac{1}{\sqrt{x}}\) can be rewritten as \(y = x^{1/2} + x^{-1/2}\). From this it is clear that the problem involves the use of the power rule for differentiation. The power rule states that the derivative of \(x^n\) with respect to \(x\) is \(n\cdot x^{n-1}\). By applying the power rule, the function \(y\) can be differentiated with respect to \(x\) to give: \(\frac{d y}{d x} = \frac{1}{2} x^{-1/2} - \frac{1}{2} x^{-3/2}\)
02

Multiply the derivative by \(2x\)

Next, the derivative of \(y\) must be multiplied by \(2x\), as the exercise states. Carefully applying the multiplication gives \(2x \cdot \frac{d y}{d x} = 2x \cdot (\frac{1}{2} x^{-1/2} - \frac{1}{2} x^{-3/2}) = x^{1/2} - x^{-1/2}\)
03

Simplify

Finally, to present the result in the form given in the exercise, the expression must be simplified. This gives \(2x \cdot \frac{d y}{d x} = \sqrt{x} - \frac{1}{\sqrt{x}}\)

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Most popular questions from this chapter

If \(\sqrt{1-x^{6}}+\sqrt{1-y^{6}}=a\left(x^{3}-y^{3}\right)\) prove that \(\frac{d y}{d x}=\frac{x^{2}}{y^{2}} \sqrt{\frac{1-y^{6}}{1-x^{6}}}\)

Find \(\frac{d^{2} y}{d x^{2}}\), if (i) \(x=a t^{2}, y=2 a t\) (ii) \(x=a \cos ^{3} \theta, y=a \sin ^{3} \theta\) (iii) \(x=a \cos \theta, y=b \sin \theta\)

If \(y=\cos ^{-1}\left\\{\frac{7}{2}(1+\cos 2 x)+\sqrt{\sin ^{2} x-48 \cos ^{2} x} \sin x\right\\}\) for all \(x\) in \(\left(0, \frac{\pi}{2}\right)\) then prove that $$ \frac{d y}{d x}=1+\frac{\sin x}{\sqrt{\sin ^{2} x-48 \cos ^{2} x}} $$

If \(x=a \cos \theta, y=b \sin \theta\), find \(\frac{d^{2} y}{d x^{2}}\)

Find \(\frac{d y}{d x}\), if \(2 x^{2}+3 x y+3 y^{2}=1\)
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