Differentiate the following function from first principles: `e^(cosx)`

Explanation

Let `f(x)=e^{cos x}`


` f(x+h)=e^{cos (x+h)}`


`therefore frac{d}{d x}{f(x)}=lim _{h -> 0} frac{f(x+h)-f(x)}{h}`


`=lim _{h -> 0} frac{e^{cos (x+h)}-e^{cos x}}{h}`


`=lim _{h -> 0} e^{cos x}[frac{e^{cos (x+h)-cos x}-1}{h}]`


`=lim _{h -> 0} e^{cos x}[frac{e^{cos (x+h)-cos x}-1}{cos (x+h) cos x}] times frac{cos (x+h)-cos x}{h}`


`=e^{cos x} lim _{h -> 0}(frac{cos (x+h)-cos x}{h}) times lim _{h -> 0}[frac{e^{cos (x+h)-cos x}-1}{cos (x+h)-cos x}]`


`=e^{cos x} lim _{h -> 0}(frac{cos (x+h)-cos x}{h})[because lim _{h -> 0} frac{e^{x}-1}{x}=1]`


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