Question

Logarithmic differentiation for


`y = x {}^(2)`

someone help me
​

Answer 1

y’ = 2x

Explanation:

Let y = f (x), take the natural logarithm of both sides ln (y) = ln (f (x))

ln (y) = ln (x²)

Differentiate the expression using the chain rule, keeping in mind that y is a function of x.

Differentiate the left hand side ln (y) using the chain rule.

y’/y = 2 In (x)

Differentiate the right hand side.

Differentiate 2 ln (x)

y’/y = d/dx = [ 2 In (x) ]

Since 2 is constant with respect to xx, the derivative of 2 ln (x) with respect to x is 2 d/dx [ln (x)]

y’/y = 2 d/dx [In (x)]

The derivative of ln (x) with respect to x is 1/x.

y’/y = 2 1/x

Combine 2 and 1/x

y’/y = 2/x

Isolate y' and substitute the original function for y in the right hand side.

y’ =
`(2)/(x)` x²

Factor x out of x².

y’ =
`(2)/(x)` (x * x)

Cancel the common factor.

y’ =
`(2)/(x)` (x * x) (The x that is under 2 and the other x that I have underlined are the ones that cancel out)

Rewrite the expression.

y’ = 2x

So therefore, the answer would be 2x.

Answer 2

`\boxed {(dy)/(dx)= 2x}`

Explanation:

Solving :

⇒ log y = log (x²)

⇒ log y = 2 log x

⇒
`\mathsf {(1)/(y) (dy)/(dx) = (1)/(x) * 2}`

⇒
`\mathsf {(dy)/(dx)= 2x}`