Question

The gradient of the curve x3+y3 = 9 at the point (1,2) is

Answer 1

Answer:

equation=x3+y3

x=1,y=2

according to problem

(1)3+(2)3=9

3+6=9

Answer 2

Answer:

The gradient of the curve $x^{3} +y^{3} =9$ at the point $(1,2)$ is $-\frac{1}{4}$

Step-by-step explanation:

Given the expression $x^{3} +y^{3} =9$ ....................................Equ(1)

 Given the point $(1,2)$ means $x=1,y=2$

 Differentiating equ(1) with respect to x we get

  $\frac{d}{dx} (x^{3} +y^{3} )=\frac{d}{dx} 9$

or  $\frac{d}{dx} x^{3} +\frac{d}{dx} y^{3} =\frac{d}{dx} 9$

or $\frac{d}{dx} x^{3} +\frac{d}{dy} y^{3} \times\frac{dy}{dx} =\frac{d}{dx} 9$

As we know $\frac{d}{dx} x^{n} =nx^{n-1}$ , n is exponent

                     $\frac{dC}{dx} =0$ , C is Constant

or $3x^{2} +3y^{2} \frac{dy}{dx} =0$

or $\frac{dy}{dx} =-\frac{3x^{2} }{3y^{2} }$

or $\frac{dy}{dx} =-\frac{x^{2} }{y^{2} }$

Now  Gradient of the curve $x^{3} +y^{3} =9$ at the point $(1,2)$

          or $\frac{dy}{dx} =-\frac{1^{2}}{2^{2}}$

     or  $\frac{dy}{dx} =-\frac{1}{4}$

gradient of the curve is $-\frac{1}{4}$