Question

If f of x = sin inverse root x minus beta by alpha minus beta and g(x)=tan^-1 root x- beta/alpha-x then prove that f'(x)=g'(x)

Answer 1

Answer:

Therefore since f(x) = g(x) we can say that f'(x) = g'(x). Hence it is proved.

Step-by-step explanation:

It is given that $f(x) = sin^{-1} \sqrt{\frac{x- \beta}{\alpha - \beta} }$   and it is also given that $g(x) = sin^{-1} \sqrt{\frac{x- \beta}{\alpha -x }$

So $f(x) = sin^{-1} \sqrt{\frac{x- \beta}{\alpha - \beta} } = cosec^{-1} \sqrt{\frac{\alpha - \beta}{ x- \beta} } = cot^{-1}\sqrt{ (\sqrt{\frac{\alpha - \beta}{ x- \beta}})^2 -1} = cot^{-1}\sqrt{ {\frac{\alpha - \beta}{ x- \beta}} -1$

Therefore $f(x) = cot^{-1}\sqrt{ {\frac{\alpha - \beta}{ x- \beta}} -1 } = cot^{-1}\sqrt{ {\frac{\alpha - x}{ x- \beta}} } = tan^{-1}\sqrt{ {\frac{x- \beta}{\alpha - x }} } = g(x)$

Therefore since f(x) = g(x) we can say that f'(x) = g'(x). Hence it is proved.

Answer 2

Answer:

Step-by-step explanation:

sorry I can't answer