Question

can anyone explain why cosxdx=d(sinx)

Answer 1

just go and Check the taylor expansions of sinx and cosx...

derivate one n ull reach the other :)

Answer 2

thanks

Answer 3

by definition the derivative of sinx is cosx
\[\lim_{h \rightarrow 0}\frac{\sin(x+h)-\sin(x)}{h}=\lim_{h \rightarrow 0}\frac{sinxcosh+sinhcosx-sinx}{h}\]
\[=\lim_{h \rightarrow 0}\frac{sinx(\cosh-1)+sinhcosx}{h}=\lim_{h \rightarrow 0}\frac{sinx(\cosh-1)}{h}+\lim_{h \rightarrow 0 }\frac{sinhsinx}{h}\]
\[=sinxlim_{h \rightarrow 0}\frac{\cosh-1}{h}+cosx \lim_{h \rightarrow 0}\frac{\sinh}{h}\]
\[=sinx*0+cosx*1=0+cosx=cosx\]

Answer 4

we could have also applied l'hospital's rule (however you spell l'hosptial)
\[\lim_{h \rightarrow 0}\frac{\sin(x+h)-\sin(x)}{h}=\lim_{h \rightarrow 0}\frac{\cos(x+h)}{1}=\cos(x+0)=cosx\]

Answer 5

i know d/dx(sinx)=cosx
but my teacher said that d/dx(sinx) x dx =cosx x dx
so d(sinx) =cosx dx
i don't know the priciple behind

Answer 6

its just treating derivatives like fractions

Answer 7

so what is the priociple

Answer 8

they are actually called"differentials" , but yeh dnt worry bout that too much

Answer 9

the principle, just treat them as fractions

Answer 10

think of dy/dx , as meaning literally "dy" divided by "dx", even though the maths professors thinks its dodgy

Answer 11

when you think about it, it makes sense that you can treat them like fractions , because a derivative is a gradient , and a gradient is a fraction

Answer 12

\[f'(x)=\frac{dy}{dx}\rightarrow dy=f'(x)dx\]

Answer 13

thx

Answer 14

You could take a look at the graphs of sin x and cosx (they are the same except that sinx is 90 degrees to the right of cos x)