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Chapter 6: Problem 134

If \(y=x \log \left(\frac{x}{a+b x}\right)\), prove that \(x^{3} \frac{d^{2} y}{d x^{2}}=\left(x \frac{d y}{d x}-y\right)^{2}\).

Short Answer

Expert verified
After applying the product rule and chain rule to differentiate, the expression simplifies and both sides of the equation \(x^{3} \frac{d^{2} y}{d x^{2}} = (x \frac{d y}{d x}-y)^{2}\) equate, confirming the given equation.

Step by step solution

01

Compute the Derivative of \(y\)

To compute the first derivative of \(y = x \log \left(\frac{x}{a+b x}\right)\), we need to apply both the product rule and the chain rule. The product rule states that the derivative of two functions \(u(x) * v(x)\) is given by \(u'(x) * v(x) + u(x) * v'(x)\) and the chain rule states that the derivative of a composed function \(f(g(x))\) is \(f'(g(x)) * g'(x)\).\nAfter applying the above rules, we find that the first derivative of \(y\) is \(\frac{d y}{d x} = \log \left(\frac{x}{a+b x}\right) + x \cdot \frac{1}{x} \cdot \frac{a}{a + bx} + x\log \left(\frac{x}{a+bx}\right) * \frac{-b x}{a + bx} = \log \left(\frac{x}{a+b x}\right) + \frac{a - b x}{a + bx}\).
02

Compute the Second Derivative of \(y\)

The second derivative of \(y\) would be the derivative of the first derivative with respect to \(x\). Again applying the chain rule to calculate the derivative, we have \(\frac{d^{2} y}{d x^{2}} = \frac{d}{dx}\left( \log \left( \frac{x}{a+bx} \right) \right) + \frac{d}{dx}\left( \frac{a - b x}{a + bx} \right)\). Evaluate these derivatives to obtain the second derivative of \(y\).
03

Simplify the Equation

Now, it is given that \(x^{3} \frac{d^{2} y}{d x^{2}}=\left(x \frac{d y}{d x}-y\right)^{2}\). Substitute the calculated values of \(\frac{d y}{d x}\) and \(\frac{d^{2} y}{d x^{2}}\) in this equation and simplify the equation to verify that both sides are equal.

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Most popular questions from this chapter

If \(y=x \sin y\), prove that \(\frac{d y}{d x}=\frac{y}{x(1-x \cos y)} .\)

If \(x=a\left(t+\frac{1}{t}\right)\) and \(y=a\left(t-\frac{1}{t}\right)\). then prove that \(\frac{d y}{d x}=\frac{x}{y}\)

If \(x^{2}-y^{2}=t-\frac{1}{t}\) and \(x^{4}+y^{4}=r^{2}+\frac{1}{t^{2}}\) then prove that \(x^{3} y \frac{d y}{d x}+1=0\)

If \(y=\left(1+\frac{1}{x}\right)^{x}+x^{\left(1+\frac{1}{x}\right)}\), find \(\frac{d y}{d x}\) at \(x=1\)

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 $$ \frac{d y}{d x}=1+\frac{\sin x}{\sqrt{\sin ^{2} x-48 \cos ^{2} x}} $$
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