Question

find the derivatives from first principle : cos²x​

Answer 1

Answer:

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Answer 2

$f'(x)=-sin2x$

Step-by-step explanation:

Let $f(x)=cos^2x$

$f(x+h)=cos^2(x+h)$

Derivative of function by  first principle

$f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

Using the formula

$f'(x)=\lim_{h\rightarrow 0}\frac{cos^2(x+h)-cos^2x}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{(cos(x+h)-cosx)(cos(x+h)+cosx)}{h}$

Using identity: $a^2-b^2=(a+b)(a-b)$

$f'(x)=\lim_{h\rightarrow 0}\frac{-2sin(\frac{2x+h}{2})sin\frac{h}{2}\times 2cos(\frac{2x+h}{2}cos\frac{h}{2}}{h}$

Using identity:

$cosx-cosy=-2sin\frac{x+y}{2}sin\frac{x-y}{2}$

$cosx+cosy=2cos\frac{x+y}{2}cos\frac{x-y}{2}$

$f'(x)=\lim_{h\rightarrow 0}\frac{-sin2(\frac{2x+h}{2})sin\frac{h}{2}cos\frac{h}{2}}{\frac{h}{2}}$

By using formula:$2sinxcosx=sin2x$

$f'(x)=\lim_{h\rightarrow 0}(-sin(2x+h))\times\lim_{h\rightarrow 0}\frac{sin\frac{h}{2}}\frac{h}{2}}$

$f'(x)=-sin2x$

By using $\lim_{h\rightarrow 0}\frac{sinx}{x}=1$

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