Derivative of Cos x (Derivative of Cosine)

What is the Derivative of Cos x ?

The derivative of cos x is -sin x.

$\frac{d}{dx}(\cos x)=-\sin x$

Proof of Derivative of Cos x

Way 1

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

$(\cos x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{\cos (x+h)-\cos x}{h}$

$\boxed{\cos (p+q)=\cos p.\cos q-\sin p.\sin q}$

$(\cos x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{\cos x.\cos h-\sin x.\sin h-\cos x}{h}$

$(\cos x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{\cos x.\cos h-\cos x-\sin x.\sin h}{h}$

$(\cos x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{\cos x.(\cos h-1)-\sin x.\sin h}{h}$

$(\cos x{)}^{\prime }=\underset{h\to 0}{\lim }[\frac{\cos x.(\cos h-1)}{h}-\frac{\sin x.\sin h}{h}]$

$(\cos x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{\cos x.(\cos h-1)}{h}-\underset{h\to 0}{\lim }\frac{\sin x.\sin h}{h}$

$(\cos x{)}^{\prime }=\cos x.\underset{h\to 0}{\lim }\frac{(\cos h-1)}{h}-\sin x.\underset{h\to 0}{\lim }\frac{\sin h}{h}$

$\boxed{\underset{t\to 0}{\lim }\frac{\sin t}{t}=1}$ $\boxed{\underset{t\to 0}{\lim }\frac{\cos t-1}{t}=0}$

$(\cos x{)}^{\prime }=\cos x.0-\sin x.1$

$(\cos x{)}^{\prime }=0-\sin x$

$(\cos x{)}^{\prime }=-\sin x$

Way 2

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

$(\cos x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{\cos (x+h)-\cos x}{h}$

$\boxed{\cos p-\cos q=-2.\sin (\frac{p-q}{2}).\sin (\frac{p+q}{2})}$

$(\cos x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{-2.\sin (\frac{\cancel{x}+h-\cancel{x}}{2}).\sin (\frac{x+h+x}{2})}{h}$

$(\cos x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{-2.\sin (\frac{h}{2}).\sin (\frac{2x+h}{2})}{h}$

$(\cos x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{-2.\sin (\frac{h}{2}).\sin (x+\frac{h}{2})}{h}$

$(\cos x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{-\sin (\frac{h}{2}).\sin (x+\frac{h}{2})}{\frac{h}{2}}$

$(\cos x{)}^{\prime }=-\underset{h\to 0}{\lim }\frac{\sin (\frac{h}{2}).\sin (x+\frac{h}{2})}{\frac{h}{2}}$

$(\cos x{)}^{\prime }=-\underset{h\to 0}{\lim }[\frac{\sin (\frac{h}{2})}{\frac{h}{2}}.\sin (x+\frac{h}{2})]$

$(\cos x{)}^{\prime }=-\underset{h\to 0}{\lim }\frac{\sin (\frac{h}{2})}{\frac{h}{2}}.\underset{h\to 0}{\lim }\sin (x+\frac{h}{2})$

$h=\frac{h}{2}$$(h\to 0)$

$(\cos x{)}^{\prime }=-\underset{h\to 0}{\lim }\frac{\sin h}{h}.\underset{h\to 0}{\lim }\sin (x+h)$

$\boxed{\underset{t\to 0}{\lim }\frac{\sin t}{t}=1}$

$(\cos x{)}^{\prime }=-1.\sin (x+0)$

$(\cos x{)}^{\prime }=-1.\sin x$

$(\cos x{)}^{\prime }=-\sin x$

Way 3

By using the infinite series expansions of the sin x and cos x functions, we can prove that the derivative of cos x is equal to -sin x. The infinite series expansions of the sin x and cos x functions are as follows.

$\boxed{\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-...}$

$\boxed{\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}-...}$

$\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}-...$

$(\cos x{)}^{\prime }=(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}-...{)}^{\prime }$

$(\cos x{)}^{\prime }=(1{)}^{\prime }-(\frac{x^2}{2!}{)}^{\prime }+(\frac{x^4}{4!}{)}^{\prime }-(\frac{x^6}{6!}{)}^{\prime }+(\frac{x^8}{8!}{)}^{\prime }-...$

$(\cos x{)}^{\prime }=0-\frac{2.x}{2!}+\frac{4.x^3}{4!}-\frac{6.x^5}{6!}+\frac{8.x^7}{8!}-...$

$(\cos x{)}^{\prime }=-\frac{\cancel{2}.x}{\cancel{2}.1!}+\frac{\cancel{4}.x^3}{\cancel{4}.3!}-\frac{\cancel{6}.x^5}{\cancel{6}.5!}+\frac{\cancel{8}.x^7}{\cancel{8}.7!}-...$

$(\cos x{)}^{\prime }=-\frac{x}{1}+\frac{x^3}{3!}-\frac{x^5}{5!}+\frac{x^7}{7!}-...$

$(\cos x{)}^{\prime }=-x+\frac{x^3}{3!}-\frac{x^5}{5!}+\frac{x^7}{7!}-...$

$(\cos x{)}^{\prime }=-(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...)$

$(\cos x{)}^{\prime }=-\sin x$

$\boxed{\Downarrow }$

$\boxed{f(x)=\cos u(x)\Rightarrow f^{\prime }(x)=-u^{\prime }(x).\sin u(x)}$ (click to see the proof of the formula)

$Question:$

$f(x)=\cos (5x^2+7x-3)\Rightarrow f^{\prime }(x)=?$

$Answer:$

$f(x)=\cos (5x^2+7x-3)$

$f^{\prime }(x)=[\cos (5x^2+7x-3){]}^{\prime }$

$\boxed{f(x)=\cos u(x)\Rightarrow f^{\prime }(x)=-u^{\prime }(x).\sin u(x)}$

$f^{\prime }(x)=-(5x^2+7x-3{)}^{\prime }.\sin (5x^2+7x-3)$

$f^{\prime }(x)=-(\mathrm{2.5.}x+7-0).\sin (5x^2+7x-3)$

$f^{\prime }(x)=-(10x+7).\sin (5x^2+7x-3)$