Derivative of Cos u(x)

What is the Derivative of Cos u(x) ?

The derivative of cos u(x) is -u'(x).sin u(x).

$\frac{d}{dx}[\cos u(x)]=-\frac{d}{dx}[u(x)].\sin u(x)$

Proof of Derivative of Cos u(x)

Way 1

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

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

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

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

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

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

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

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

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

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

$h=\frac{u(x+h)-u(x)}{2}$$(h\to 0)$

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

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

$[\cos u(x){]}^{\prime }=-1.u^{\prime }(x).\sin [\frac{u(x+0)+u(x)}{2}]$

$[\cos u(x){]}^{\prime }=-1.u^{\prime }(x).\sin [\frac{u(x)+u(x)}{2}]$

$[\cos u(x){]}^{\prime }=-1.u^{\prime }(x).\sin [\frac{\cancel{2}.u(x)}{\cancel{2}}]$

$[\cos u(x){]}^{\prime }=-u^{\prime }(x).\sin u(x)$

Way 2

By making use of the infinite series expansions of the sin u(x) and cos u(x) functions, we can prove that the derivative of cos u(x) is equal to -u'(x).sin u(x). The infinite series expansions of sin u(x) and cos u(x) functions are as follows.

$\boxed{\sin u(x)=u(x)-\frac{[u(x){]}^3}{3!}+\frac{[u(x){]}^5}{5!}-\frac{[u(x){]}^7}{7!}+\frac{[u(x){]}^9}{9!}-...}$

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

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

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

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

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

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

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

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

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

$[\cos u(x){]}^{\prime }=-u^{\prime }(x).\sin u(x)$

$Question:$

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

$Answer:$

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

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

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

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

$f^{\prime }(x)=-(\mathrm{3.3.}{x}^2-\mathrm{2.2.}x).\sin (3x^3-2x^2)$

$f^{\prime }(x)=-(9x^2-4x).\sin (3x^3-2x^2)$