Given-that-f-sin-x-cos-x-evaluate-f-sin-45-

Question Number 142844 by ZiYangLee last updated on 06/Jun/21

Given that f (\sin x) = \cos x, evaluate

f^{'} (\sin 45 °) .

Answered by meetbhavsar25 last updated on 06/Jun/21

f (\sin x) = \cos x = \sqrt{1 - \sin^{2} x}

s o

f (x) = \sqrt{1 - x^{2}}

f^{'} (x) = \frac{1}{2 \sqrt{1 - x^{2}}} \times (- 2 x)

f^{'} (x) = \frac{- x}{\sqrt{1 - x^{2}}}

f^{'} (\sin 45 °) = f^{'} (\frac{1}{\sqrt{2}}) = \frac{- \frac{1}{\sqrt{2}}}{\sqrt{1 - \frac{1}{2}}} = \frac{- \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} = - 1

S o, a n s w e r i s - 1 .

Answered by Dwaipayan Shikari last updated on 06/Jun/21

f (s i n x) = c o s x

f^{'} (s i n x) c o s x = - s i n x

f^{'} (s i n x) = - t a n x \Rightarrow f^{'} (s i n 45 °) = - 1