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Fine The Gradient of the curve xy=6, at point 1 & 6.
[tex] \\ \\ \\ \\ \\ \\ \\ [/tex]
Thnx ~~​

Sagot :

Ankit

Answer:

[tex]\fbox{Gradient of curve = -6}[/tex]

Step by step explanation:

Given:

Equation of curve ↦ xy = 6

Point ↦(1,6)

To find:

The gradient of curve = ?

Solution:

[tex]Using \: the \: equation, \\ xy = 6 \\ [/tex]

Taking log both the side,

[tex]log(xy) = log(6) \\ log(x) + log(y) = log(6) \\ Taking \: derivative \: both \: the \: side \\ \frac{1}{x} + \frac{1}{y} \cdot\frac{dy}{dx} = \: 0 \\ (Derivative \: of \: any \: constant =0) \\ \frac{1}{y} \cdot\frac{dy}{dx} = - \frac{1}{x} \\ \frac{dy}{dx} = - \frac{y}{x} \\ To \: find \: gradient \: at \: point (1,6) \\ \frac{dy}{dx} \mid_{(1,6)} = \frac{ - 6}{1} \\ \frac{dy}{dx} \mid_{(1,6)} = - 6 \\ \fbox{Gradient of curve = -6}[/tex]

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