Derivative of Cos u

What is the Derivative of Cos u ?

The derivative of cos u is -u'.sin u.

$(cosu{)}^{\prime }=-u^{\prime }.sinu$

$\frac{d}{dx}(cosu)=-u^{\prime }.sinu$

Proof of the Derivative of Cos u

Way 1

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

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

$\boxed{cosp-cosq=-2.sin\frac{p+q}{2}.sin\frac{p-q}{2}}$

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

$[cosu(x){]}^{\prime }=\underset{h\to 0}{\lim }\frac{-sin\frac{u(x+h)+u(x)}{2}.sin\frac{u(x+h)-u(x)}{2}}{h.\frac{1}{2}}$

$[cosu(x){]}^{\prime }=\underset{h\to 0}{\lim }\frac{-[u(x+h)-u(x)].sin\frac{u(x+h)+u(x)}{2}.sin\frac{u(x+h)-u(x)}{2}}{h.\frac{1}{2}.[u(x+h)-u(x)]}$

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

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

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

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

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

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

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

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

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

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

$[cosu(x){]}^{\prime }=-u^{\prime }(x).1.sinu(x)$

$[cosu(x){]}^{\prime }=-u^{\prime }(x).sinu(x)$

$u(x)=u$

$u^{\prime }(x)=u^{\prime }$

$(cosu{)}^{\prime }=-u^{\prime }.sinu$

Way 2

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

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

$u(x)=u$

$u^{\prime }(x)=u^{\prime }$

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

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

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

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

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

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

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

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

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

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

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

$(cosu{)}^{\prime }=-u^{\prime }.sinu$

Question

$f(x)=cos(-2x^3+4x)\Rightarrow f^{\prime }(x)=?$

Answer

$(cosu{)}^{\prime }=-u^{\prime }.sinu$

$[cos(-2x^3+4x){]}^{\prime }=-(-2x^3+4x{)}^{\prime }.sin(-2x^3+4x)$

$[cos(-2x^3+4x){]}^{\prime }=-(-\mathrm{2.3.}{x}^2+4).sin(-2x^3+4x)$

$[cos(-2x^3+4x){]}^{\prime }=-(-6x^2+4).sin(-2x^3+4x)$

$[cos(-2x^3+4x){]}^{\prime }=(6x^2-4).sin(-2x^3+4x)$