Question:

Find the derivative of cos x from first principle.

Updated On: Oct 25, 2023
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Solution and Explanation

Let f (x) = cos x. Accordingly, from the first principle,
f'(x)=limh0\lim_{h\rightarrow 0} f(x+h)f(x)h\frac{f(x+h)-f(x)}{h}
=limh0\lim_{h\rightarrow 0} cos(x+h)cos(x)h\frac{cos(x+h)-cos(x)}{h}
=limh0\lim_{h\rightarrow 0} [cosxcoshsinxsinhcosxh\frac{cos\,xcos\,h-sin\,xsin\,h-cos\,x}{h}]
=limh0\lim_{h\rightarrow 0} [cosx(1cosh)hsinxsinhh\frac{-cos\,x(1-cos\,h)}{h}-\frac{sin\,xsin\,h}{h}]
=-cos(limh0\lim_{h\rightarrow 0} 1coshh\frac{1-cos\,h}{h}) - sinx limh0\lim_{h\rightarrow 0}(sinhh\frac{sin\,h}{h})
=-cosx(0 ) - sinx(1) [limh0\lim_{h\rightarrow 0} 1coshh\frac{1-cos\,h}{h} =0 and limh0\lim_{h\rightarrow 0} sinhh\frac{sin\,h}{h} = 1]
=-sinx
∴f'(x) = -sinx
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