Question

Find derivative of f(x) = square root of cosx using first principles.

Answer 1

Answer:

f'(x) = - sin x / (2√cos x)

Step-by-step explanation:

f(x) = √cos x

f(x) = (cos x)^½

f'(x) = (½) (cos x) ^-½ (-sin x)

f'(x) = - sin x / (2√cos x)

Answer 2

GIVEN:- f(x) = √cosx

we know by first principles f'(x) = lim_(h->0) {f(x+h) - f(x)}/h

so,

$\sf{f'(x) = lim_{h \rightarrow0} \frac{ \sqrt{cos(x + h)} - \sqrt{cosx}}{h}}$

$\sf{f'(x) = lim_{h \rightarrow0} \frac{ \sqrt{cos(x + h)} - \sqrt{cosx}}{h} \times \frac{\sqrt{cos(x+h)}+\sqrt{cosx}}{\sqrt{cos(x+h)}+\sqrt{cosx}}}$

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

$\sf{f'(x) = lim_{h\rightarrow0}\frac{cosxcosh - sinxsinh - cosx}{h\{\sqrt{cos(x+h)}+\sqrt{cosx}\}}}$

$\sf{f'(x) = lim_{h\rightarrow0}\frac{-cosx(1-cosh) - sinxsinh}{h\{\sqrt{cos(x+h)}+\sqrt{cosx}\}}}$

$\sf{f'(x) = \frac{-cosx\:{lim_{h\rightarrow0}\frac{(1-cosh)}{h}}-sinx\:{lim_{h\rightarrow0}\frac{sinh}{h}}}{lim_{h\rightarrow0}\{\sqrt{cos(x+h)}+\sqrt{cosx}\}}}$

we know:- $\bold{lim_{h\rightarrow0}\frac{(1-cosh)}{h}=0 \:\:\: and \:\:lim_{h\rightarrow0} \frac{sinh}{h}=1}$

$\sf{f'(x) = \frac{-cosx(0) - sinx(1)}{\{\sqrt{cos(x+0)}+\sqrt{cosx}\}}}$

$\sf{f'(x) = \frac{-sinx}{2\sqrt{cosx}}}$