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Chapter 6: Problem 11

If \(y=\sqrt{x}+x \sqrt{x}+x^{2} \sqrt{x}+x^{3} \sqrt{x}\), find \(\frac{d y}{d x}\)

Short Answer

Expert verified
\(\frac{d y}{d x} = 0.5x^{-0.5} + \sqrt{x} + 0.5x^{0.5} + 2x\sqrt{x} + 0.5x^{1.5} + 3x^{2}\sqrt{x} + 0.5x^{2.5}\)

Step by step solution

01

Derive first term

The first term which is \(\sqrt{x}\) is simple, its derivative will be \(0.5x^{-0.5}\).
02

Differentiate the second term

The second term is \(x \sqrt{x}\) which is a product of two functions: \(x\) and \(\sqrt{x}\). Therefore, use the product rule to differentiate: \((fg)' = f'g + fg'\) where \(f = x\) and \(g = \sqrt{x}\). Thus, the derivative will be \(1*\sqrt{x} + x*0.5x^{-0.5} = \sqrt{x} + 0.5x^{0.5}\).
03

Differentiate the third term

The third term \(x^{2} \sqrt{x}\) is also a product of two functions, \(x^{2}\) and \(\sqrt{x}\). Apply the product rule again which will give us \(2x*\sqrt{x} + x^{2}*0.5x^{-0.5} = 2x\sqrt{x} + 0.5x^{1.5}\).
04

Differentiate the fourth term

The last term is \(x^{3} \sqrt{x}\), which is a product of two functions, \(x^{3}\) and \(\sqrt{x}\). Again, apply the product rule which will give us \(3x^{2}*\sqrt{x} + x^{3}*0.5x^{-0.5} = 3x^{2}\sqrt{x} + 0.5x^{2.5}\).
05

Sum all partial derivatives to get \(\frac{d y}{d x}\)

From the previous steps, we derived each term separately. Now, to get the derivative of the whole function \(y\), sum up all these derivatives. The answer is: \(0.5x^{-0.5} + \sqrt{x} + 0.5x^{0.5} + 2x\sqrt{x} + 0.5x^{1.5} + 3x^{2}\sqrt{x} + 0.5x^{2.5}\)

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Most popular questions from this chapter

If \(e^{x}+e^{y}=e^{x+y}\), prove that, \(\frac{d y}{d x}+e^{y-x}=0\)

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