Question

Differentiate the following functions from first principles e^cos x.

Answer 1

\[\text{Let } f \left( x \right) = e^{\cos x} \]
\[ \Rightarrow f\left( x + h \right) = e^{\cos\left( x + h \right)} \]
\[\therefore \frac{d}{dx}\left\{ f\left( x \right) \right\} = \lim_{h \to 0} \frac{f\left( x + h \right) - f\left( x \right)}{h}\]
\[ = \lim_{h \to 0} \frac{e^{\cos\left( x + h \right)} - e^{\cos x}}{h}\]
\[ = \lim_{h \to 0} e^{\cos x }\left[ \frac{e^{\cos\left( x + h \right) - \cos x} - 1}{h} \right]\]
\[ = \lim_{h \to 0} e^{\cos x} \left[ \frac{e^{\cos\left( x + h \right) - \cos x} - 1}{\cos\left( x + h \right) \cos x} \right] \times \frac{\cos\left( x + h \right) - \ cosx}{h}\]
\[ = e^{\cos x} \lim_{h \to 0} \left( \frac{cos\left( x + h \right) - \cos x}{h} \right) \times \lim_{h \to 0} \left[ \frac{e^{\cos\left( x + h \right) - \ cos x} - 1}{\cos\left( x + h \right) - \cos x} \right]\]
\[ = e^{\cos x} \lim_{h \to 0} \left( \frac{\cos\left( x + h \right) - \cos x}{h} \right) \left[ \because \lim_{h \to 0} \frac{e^x - 1}{x} = 1 \right]\]
\[ = e^{\cos x} \lim_{h \to 0} \left\{ \frac{- 2\sin\left( \frac{x + h + x}{2} \right)\sin\left( \frac{x + h - x}{2} \right)}{h} \right\} \left[ \because \cos A - \cos B = - 2\sin \left( \frac{A + B}{2} \right)\sin\left( \frac{A - B}{2} \right) \right]\]
\[ = e^{\cos x} \lim_{h \to 0} \frac{- \sin\left( \frac{2x + h}{2} \right)}{1} \times \frac{\sin\left( \frac{h}{2} \right)}{\frac{h}{2}}\]
\[ = e^{\cos x} \lim_{h \to 0} \frac{- \sin\left( \frac{2x + h}{2} \right)}{1} \times \lim_{h \to 0} \frac{\sin\left( \frac{h}{2} \right)}{\frac{h}{2}}\]
\[ = e^{\cos x} \lim_{h \to 0} - \sin\left( \frac{2x + h}{2} \right) \left[ \because \frac{\sin x}{x} = 1 \right]\]
\[ = e^{\cos x} \left( - \sin x \right)\]
\[ = - \sin x e^{\cos x} \]
\[\text{ Hence }, \frac{d}{dx}\left( e^{\cos x} \right) = - \sin x e^{\cos x }\]