Question

if x³−y²=1 find dy/dx​

Answer 1

Answer:

Step-by-step explanation:

Given, x  

2

+y  

2

=1  

Differentiating w.r.t x,  

2x+2yy  

′

=0

⇒x+yy  

′

=0

Again differentiating w.r.t. x, we get

1+yy  

′′

+y  

′

y  

′

=0

1+yy  

′′

+y  

′2

=0

Answer 2

$\large\sf\underline{\underline{ \red{Question}}} :$
⠀⠀ ⠀ ⠀● if x³−y²=1 , Find dy/dx​

$\large\sf\underline{\underline{ \red{Solution}}} :$

We know that,
$\textsf{Derivative of x^n} =\sf n.x^{n-1}$

So,
$\textsf{ Diferentiating both sides w.r.t x,}$
$\implies \sf 3x^2 - 2y.\frac{dy}{dx} = 0\\\\\sf\implies - 2y.\frac{dy}{dx} = -3x^2 \\\\\implies\sf \dfrac{dy}{dx} = \dfrac{\cancel{-}3x^2}{\cancel{-}2y} \\\\\implies\sf \dfrac{dy}{dx} =\pink{ \dfrac{3x^2}{2y}}$

$\large\bf Hence,$
Differentiation of x³-y²= 1 w.r.t x is 3x²/2y .