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Sholpan [36]
2 years ago
11

Logarithmic differentiation for

le="y = x {}^{2} " alt="y = x {}^{2} " align="absmiddle" class="latex-formula">

someone help me
​
Mathematics
2 answers:
slavikrds [6]2 years ago
6 0

Answer:

y’ = 2x

Step-by-step 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’ = \frac{2}{x} x²

Factor x out of x².

y’ = \frac{2}{x} (x * x)

Cancel the common factor.

y’ = \frac{2}{x} (<u>x</u> * 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.

kirill115 [55]2 years ago
5 0

Answer:

\boxed {\frac{dy}{dx}= 2x}

Step-by-step explanation:

<u>Solving</u> :

⇒ log y = log (x²)

⇒ log y = 2 log x

⇒ \mathsf {\frac{1}{y} \frac{dy}{dx} = \frac{1}{x} \times 2}

⇒ \mathsf {\frac{dy}{dx}= 2x}

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