Question

differenation of √x2-2ax/a2-2ab

Answer 1

Step-by-step explanation:

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Answer 2

Question: Find the differentiation of the function: $\frac{\sqrt{x^2-2ax} }{a^2-2ab}$.

Answer:

The value of the differentiation of the function $\frac{\sqrt{x^2-2ax} }{a^2-2ab}$ is $\frac{x-a}{(a^2-2ab)\sqrt{x^2-2ax} }$.

Step-by-step explanation:

Differentiation is the process rate of change in function with respect to the specified variable

Recall the Chain Rule,

$\frac{d}{dx}f(g(x)) = f'(g(x))\cdot g'(x)$, where $f$ and $g$ are differentiable functions.

Consider the function as follows:

$y=\frac{\sqrt{x^2-2ax} }{a^2-2ab}$

Differentiate both the sides with respect to $x$ as follows:

$\frac{dy}{dx}= \frac{d}{dx}\left(\frac{\sqrt{x^2-2ax} }{a^2-2ab} \right)$

Notice that the function in the denominator is independent of $x$ or a constant function. So, take the function $(a^2-2ab)$ out.

$\frac{dy}{dx}=\frac{1}{{a^2-2ab}} \left[ \frac{d}{dx}\left(\sqrt{x^2-2ax} } \right)\right]$

Using chain rule, differentiate the function as follows:

$\frac{dy}{dx}=\frac{1}{{a^2-2ab}} \left[ \left(\frac{1}{2\sqrt{x^2-2ax} } \right)\cdot\frac{d}{dx}(x^2-2ax) \right]$ (Since $\frac{d}{dx}(\sqrt{x})=\frac{1}{2\sqrt{x} }$)

$\frac{dy}{dx}=\frac{1}{{a^2-2ab}}\left(\frac{1}{2\sqrt{x^2-2ax} } \right) \cdot\frac{d}{dx}(x^2-2ax)$

Further, simplify as follows:

$\frac{dy}{dx}=\frac{1}{{a^2-2ab}}\left(\frac{1}{2\sqrt{x^2-2ax} } \right) \cdot(2x-2a)$ (Since $\frac{d}{dx}(x^n)=x^{n-1}$)

$\frac{dy}{dx}=\frac{2(x-a)}{2(a^2-2ab)\sqrt{x^2-2ax} }$

$\frac{dy}{dx}=\frac{x-a}{(a^2-2ab)\sqrt{x^2-2ax} }$

Therefore, the value of the differentiation of the function $\frac{\sqrt{x^2-2ax} }{a^2-2ab}$ is $\frac{x-a}{(a^2-2ab)\sqrt{x^2-2ax} }$.

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