Question

Find the derivative of the following function:

5sin x – 6cos x + 7

Answer 1

Let f (x) = 5sin x – 6cos x + 7. Accordingly, from the first principle,

${\displaystyle f\prime \left(x\right)}$ = ${\displaystyle \underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}}$

= ${\displaystyle \underset{h\to 0}{lim}\frac{1}{h}[5\sin (x+h)-6\cos (x+h)+7-5\sin x+6\cos x-7]}$

= ${\displaystyle \underset{h\to 0}{lim}\frac{1}{h}[5\{\sin (x+h)-\sin x\}-6\{\cos (x+h)-\cos x\}]}$

= ${\displaystyle 5\underset{h\to 0}{lim}\frac{1}{h}[\sin (x+h)-\sin x]-6\underset{h\to 0}{lim}\frac{1}{h}[\cos (x+h)-\cos x]}$

= ${\displaystyle 5\underset{h\to 0}{lim}\frac{1}{h}\left[2\cos \left(\frac{x+h+x}{2}\right)\sin \left(\frac{x+h-x}{2}\right)\right]-6\underset{h\to 0}{lim}\frac{\cos x\cos h-\sin x\sin h-\cos x}{h}}$

= ${\displaystyle 5\underset{h\to 0}{lim}\frac{1}{h}\left[2\cos \left(\frac{2x+h}{h}\right)\sin \frac{h}{2}\right]-6\underset{h\to 0}{lim}\left[\frac{-\cos x(1-\cos h)-\sin x\sin h}{h}\right]}$

= ${\displaystyle 5\underset{h\to 0}{lim}\left(\cos \left(\frac{2x+h}{2}\right)\frac{\sin \frac{h}{2}}{\frac{h}{2}}\right)-6\underset{h\to 0}{lim}[\frac{-\cos x(1-\cos h)}{h}-\frac{\sin x\sin h}{h}]}$

= ${\displaystyle 5\left[\underset{h\to 0}{lim}\left(\frac{2x+h}{2}\right)\right]\left[\underset{h\to 0}{lim}\frac{\sin \frac{h}{2}}{\frac{h}{2}}\right]-6[(-\cos x)\left(\underset{h\to 0}{lim}\frac{1-\cos h}{h}\right)-\sin x\underset{h\to 0}{lim}\left(\frac{\sin h}{h}\right)]}$

= 5 cosx. 1 - 6 [(-cos x) (0) -sinx.1]

= 5 cosx + 6 sin x