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Chapter 6: Problem 10

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

Short Answer

Expert verified
-x^{-2}-4*x^{-3}+9*x^{-4}

Step by step solution

01

Rewriting the function y

Rewrite the terms in the given function in the form \(x^n\) to make them amenable to the power rule. So, \(y=x^{-1}+2*x^{-2}+3*x^{-3}\)
02

Apply the Power Rule

Applying the power rule \(d/dx[x^n]=n*x^{n-1}\), the derivative of each term of y with respect to x is calculated as follows: \n\n - \(d/dx[x^{-1}]= -1*x^{-2}\), \n\n - \(d/dx[2*x^{-2}]= -4*x^{-3}\), \n\n - \(d/dx[3*x^{-3}]= 9*x^{-4}\)
03

Simplify the Resultant Derivative

Combine all the derived terms for the complete derivative of y. Therefore, \(\frac{d y}{d x} = -x^{-2}-4*x^{-3}+9*x^{-4}\)

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Most popular questions from this chapter

Find \(\frac{d y}{d x}\), if \(y=x^{\sin x}\)

If \(y=e^{\tan ^{-1}} x\), show that $$ \left(x^{2}+1\right) \frac{d^{2} y}{d x^{2}}+(2 x-1) \frac{d y}{d x}=0 $$

If \(y=\frac{a x+b}{A x+B}\) and \(z=\frac{a y+b}{A y+B}\), prove that $$ \frac{y^{\prime \prime \prime}}{y^{\prime}}-\frac{3}{2}\left(\frac{y^{\prime \prime}}{y^{\prime}}\right)^{2}=0=\frac{z^{\prime \prime \prime}}{z^{\prime}}-\frac{3}{2}\left(\frac{z^{\prime \prime}}{z^{\prime}}\right)^{2} $$

If \(f(x)=a x+b, x \in[-2,2]\), then the point \(c \in(-2,2)\) where $$ f(c)=\frac{f(2)-f(-2)}{4} $$ (a) does not exist (b) can be any \(c \in(-2,2)\) (c) can be only 1 (d) can be only \(-1\).

If \(\sqrt{x^{2}+y^{2}}=a e^{\tan ^{-1}} x\), where \(a>0, y(0) \neq 0\) then find the value of \(y^{\prime \prime}(0)\).
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