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Let x^{k}+y^{k}=a^{k},(a,k> 0)  and  \frac{dy}{dx}+\left [ \frac{y}{x} \right ]^{\frac{1}{3}}=0, then k is :
Option: 1 \frac{1}{3}
Option: 2 \frac{3}{2}
Option: 3 \frac{2}{3}
Option: 4 \frac{4}{3}
 

Answers (1)

best_answer

 

 

Derivative of the Polynomial Function -

Derivative of the Polynomial Function

Finding derivatives of functions by using the definition of the derivative can be a lengthy and, for certain functions, a rather challenging process.

In this section, we will learn rules for finding derivatives that allow us to bypass this process. Let’s begin with the basics.

\\\mathbf{1.}\;\;\;\;\mathrm{\mathbf{\frac{\mathit{d}}{\mathit{dx}}(constant)=0}}

\\\mathbf{2.}\;\;\;\;\mathrm{\mathbf{\frac{\mathit{d}}{\mathit{dx}}(x^n)=nx^{n-1}}}

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\\x^k+y^k=a^k\;\;\;\;\;\;\;(a,k>0)\\\frac{dy}{dx}+\left (\frac{y}{x} \right )^{1/3}=0\\\Rightarrow \;kx^{k-1}+ky^{k-1}\cdot\frac{dy}{dx}=0\\\frac{dy}{dx}+\left (\frac{x}{y} \right )^{k-1}=0\\k-1=-1/3\\k=2/3

Correct Option (3)

Posted by

Kuldeep Maurya

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