Question

If y=x log (x^2-3) , then dy/dx = ?​

Answer 1

Answer:

2x^2/(x^2 -3) + log (x^2 - 3)

Step-by-step explanation:

y = x log (x^2 -3)

Differentiating both sides w.r.t. x

dy/dx = x* d/dx log (x^2 -3) + log (x^2 -3)* d/dx x ( since u*v = u d/dx v + v d/dx u)

dy/dx = x* 1/(x^2 -3)* d/dx (x^2 - 3) + log (x^2 - 3)*1

dy/dx = x/(x^2 - 3) * (2x) + log (x^2 - 3)

dy/dx = 2x^2/(x^2 -3) + log (x^2 - 3)

Answer 2

Concept Introduction:

The area of mathematics concerned with determining the existence and characteristics of function derivatives and integrals using techniques that were first based on the addition of minute differences.

Given, $y=x log (x^2-3)$

To find, $dy/dx$

Solution:

$dy/dx = x* d/dx log (x^2 -3) + log (x^2 -3)* d/dx x ( since u*v = u d/dx v + v d/dx u)dy/dx = x* 1/(x^2 -3)* d/dx (x^2 - 3) + log (x^2 - 3)*1dy/dx = x/(x^2 - 3) * (2x) + log (x^2 - 3)dy/dx = 2x^2/(x^2 -3) + log (x^2 - 3)$

Final Answer:

The final answer is $2x^2/(x^2 -3) + log (x^2 - 3)$

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