Derivative of Sin u

What is the Derivative of Sin u ?

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

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

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

Proof of the Derivative of Sin u

Way 1

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

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

$\boxed{sinp-sinq=2.sin\frac{p-q}{2}.cos\frac{p+q}{2}}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$u(x)=u$

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

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

Way 2

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

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

$u(x)=u$

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

$sinu=u-\frac{u^3}{3!}+\frac{u^5}{5!}-\frac{u^7}{7!}+\frac{u^9}{9!}-...$

$(sinu{)}^{\prime }=(u-\frac{u^3}{3!}+\frac{u^5}{5!}-\frac{u^7}{7!}+\frac{u^9}{9!}-...{)}^{\prime }$

$(sinu{)}^{\prime }=u^{\prime }-(\frac{u^3}{3!}{)}^{\prime }+(\frac{u^5}{5!}{)}^{\prime }-(\frac{u^7}{7!}{)}^{\prime }+(\frac{u^9}{9!}{)}^{\prime }-...$

$(sinu{)}^{\prime }=u^{\prime }-\frac{(u^3{)}^{\prime }}{3!}+\frac{(u^5{)}^{\prime }}{5!}-\frac{(u^7{)}^{\prime }}{7!}+\frac{(u^9{)}^{\prime }}{9!}-...$

$(sinu{)}^{\prime }=u^{\prime }-\frac{3.u^2.u^{\prime }}{3!}+\frac{5.u^4.u^{\prime }}{5!}-\frac{7.u^6.u^{\prime }}{7!}+\frac{9.u^8.u^{\prime }}{9!}-...$

$(sinu{)}^{\prime }=u^{\prime }-\frac{\cancel{3}.u^2.u^{\prime }}{\cancel{3}.2!}+\frac{\cancel{5}.u^4.u^{\prime }}{\cancel{5}.4!}-\frac{\cancel{7}.u^6.u^{\prime }}{\cancel{7}.6!}+\frac{\cancel{9}.u^8.u^{\prime }}{\cancel{9}.8!}-...$

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

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

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

Question

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

Answer

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

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

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