Question

Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed nonzero constants and m and n are integers): \(\frac{sin(x+a)}{cos\,a}\)

Answer 1

Let f(x)= \(\frac{sin(x+a)}{cos\,a}\)
By the quotient rule,
f'(x)=\(\frac{cos\,x\frac{d}{dx}[sin(x+a)]-sin(x+a)\frac{d}{dx}cos\,x}{cos^2x}\)
f'(x)=\(\frac{cos\,x\frac{d}{dx}[sin(x+a)]-sin(x+a)(-sin\,x)}{cos^2x}\) ...(i)
Let g(x)=sin(x+a). Accordingly, g(x+h)=sin(x+h+a)
By first principle,
g'(x)= \(\lim_{h\rightarrow 0}\) \(\frac{g(x+h)-g(x)}{h}\)
= \(\lim_{h\rightarrow 0}\) \(\frac{1}{h}\) [sin(x+h+a) - sin(x+a)]
= \(\lim_{h\rightarrow 0}\) \(\frac{1}{h}\)[ 2cos(\(\frac{(2x+2h+h)}{2}\)) sin(\(\frac{h}{2}\))]
= \(\lim_{h\rightarrow 0}\) [cos(\(\frac{(2x+2h+h)}{2}\))\(\frac{sin(\frac{h}{2})}{(\frac{h}{2})}\)]
=cos\(\frac{(2x+2a)}{2}\) \(\times\)1
= cos(x+a) ...(ii)
From (i) and (ii), we obtain
f'(x)= \(\frac{cos\,x.cos(x+a)+sin\,xsin(x+a)}{cos^2x}\)
=\(\frac{cos(x+a-x)}{cos^2x}\)
=\(\frac{cos\,a}{cos^2\,x}\)