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

$$\mathrm{Given}\mathrm{that}f\left(\sin x\right)=\cos x,\mathrm{evaluate}$$$$f{}^{\prime }\left(\sin 45°\right).$$

Answered by meetbhavsar25 last updated on 06/Jun/21

$$$$$$f\left(\sin x\right)=\cos x=\sqrt{1-\sin {}^{2}x}$$$$so$$$$f\left(x\right)=\sqrt{1-x^2}$$$$f^{\prime }\left(x\right)=\frac{1}{2\sqrt{1-x^2}}\times (-2x)$$$$f^{\prime }\left(x\right)=\frac{-x}{\sqrt{1-x^2}}$$$$f^{\prime }\left(\sin 45°\right)=f^{\prime }\left(\frac{1}{\sqrt{2}}\right)=\frac{-\frac{1}{\sqrt{2}}}{\sqrt{1-\frac{1}{2}}}=\frac{-\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}=-1$$$$So,answeris-1.$$

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

$$f\left(sinx\right)=cosx$$$$f^{\prime }\left(sinx\right)cosx=-sinx$$$$f^{\prime }\left(sinx\right)=-tanx\Rightarrow f^{\prime }\left(sin45°\right)=-1$$

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