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yKpoI14uk [10]
4 years ago
15

Find the derivative

Mathematics
1 answer:
hoa [83]4 years ago
3 0

This is loaded question, but let me take a stab at it:

\frac{d}{dx} \cot^{-1} \sqrt{x-1} = \frac{d\cot u}{du}\cdot \frac{du}{dx}

with u=\sqrt{x-1}

Start with the first part of the chain rule term:

y = \cot^{-1} u\\\frac{dy}{du} = \frac{d \cot^{-1}u}{du}

because cot^-1 is inverse to cot we also know that

\cot y = u

and use it in the earlier equation:

\frac{dy}{du} = \frac{d \cot^{-1}u}{du}\\\frac{1}{\frac{d\cot y}{dy}} = \frac{d \cot^{-1}u}{du}\\-\sin^2y=\frac{d \cot^{-1}u}{du}\\-\frac{1}{\cot^2 y +1} = \frac{d \cot^{-1}u}{du}\\-\frac{1}{u^2 +1} = \frac{d \cot^{-1}u}{du}

Now, the second term of the chain rule:

\frac{du}{dx}=\frac{d\sqrt{x-1}}{dx} = \frac{1}{2\sqrt{x-1}}

Putting it together:

\frac{d \cot^{-1}u}{du}\cdot\frac{du}{dx} = -\frac{1}{x-1 +1}\cdot\frac{1}{2\sqrt{x-1}} = - \frac{1}{2x\sqrt{x-1}}

and that is the final derivative of your expression.

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