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find using first principles the derivative of cosx

Guest Nov 26, 2014

Best Answer 

 #1
avatar+26399 
+10

d(cosx)/dx

.

Alan  Nov 26, 2014
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3+0 Answers

 #1
avatar+26399 
+10
Best Answer

d(cosx)/dx

.

Alan  Nov 26, 2014
 #2
avatar+91435 
+5

$$\\\displaystyle \lim_{h\rightarrow 0}\;\;\;\frac{f(x+h)-(x)}{h}\\\\
=\displaystyle \lim_{h\rightarrow 0}\;\;\;\frac{cos(x+h)-cos(x)}{h}\\\\
=\displaystyle \lim_{h\rightarrow 0}\;\;\;\frac{cosx*cosh-sinx*sinh-cos(x)}{h}\\\\
=\displaystyle \lim_{h\rightarrow 0}\;\;\;\frac{cosx(cosh-1)-sinx*sinh}{h}\\\\
=\displaystyle \lim_{h\rightarrow 0}\;\;\;\frac{cosx(cosh-1)}{h}-\frac{sinx*sinh}{h}\\\\
=\displaystyle \lim_{h\rightarrow 0}\;\;\;cosx*\frac{(cosh-1)}{h}-sinx*\frac{sinh}{h}\\\\
=cosx*0-sinx*1\\\\
=0-sin(x)\\\\
=-sin(x)$$

 

If you need these proved that also be done

 

$$\\\displaystyle \lim_{h\rightarrow 0}\;\;\;\frac{cosh-1}{h}=0\\\\
\displaystyle \lim_{h\rightarrow 0}\;\;\;\frac{sinh}{h}=1\\\\$$

Melody  Nov 26, 2014
 #3
avatar+18827 
+5

find using first principles the derivative of cosx

$$\boxed{
e^{i\phi}= \cos{(\phi)}+i\sin{(\phi)}
}$$

$$e^{i\phi}= \cos{(\phi)}+i\sin{(\phi)} \quad | \quad \frac {d}{dx}$$

$$\left(
e^{i\phi}
\right)'
= ie^{i\phi}
=
\left( \cos{(\phi)}+i\sin{(\phi)} \right) ' = (\cos{(\phi)})'+i(\sin{(\phi)})'$$

$$ie^{i\phi}
= (\cos{(\phi)})'+i(\sin{(\phi)})' \\
i\left( \cos{(\phi)}+i\sin{(\phi)}
\right) = (\cos{(\phi)})'+i(\sin{(\phi)})' \\
i\cos{(\phi)}+\underbrace{i^2}_{i^2=-1}\sin{(\phi)} = (\cos{(\phi)})'+i(\sin{(\phi)})' \\
i\cos{(\phi)}-\sin{(\phi)} = (\cos{(\phi)})'+i(\sin{(\phi)})' \\
\textcolor[rgb]{1,0,0}{ -\sin{(\phi)} } + i\textcolor[rgb]{0,0,1}{\cos{(\phi)} }= \textcolor[rgb]{1,0,0}{ (\cos{(\phi)})'}+i\textcolor[rgb]{0,0,1} {(\sin{(\phi)})'}\\ \\
\textcolor[rgb]{1,0,0}{\boxed{ (\cos{(\phi)})'= -\sin{(\phi)} }} \\
\textcolor[rgb]{0,0,1}{\boxed{ (\sin{(\phi)})'= \cos{(\phi)} }} \\$$

heureka  Nov 26, 2014

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