# Use Definition to find Derivative

• Oct 14th 2009, 02:33 PM
VitaX
Use Definition to find Derivative
Find the derivative of $Secx$ using the definition below:

$\frac{d}{dx}(x) = \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h}$

Note $y=Secx \rightarrow Cos^{-1}x$

$\frac{d}{dx}(Secx) = \lim_{h \rightarrow 0} \frac{Cos^{-1}(x + h) - Cos^{-1}(x)}{h}$

$= \lim_{h \rightarrow 0} \frac{1}{h}\left(\frac{1}{Cos(x + h)} - \frac{1}{Cos(x)}\right)$

$= \lim_{h \rightarrow 0} \frac{1}{h}\left(\frac{Cos(x) - Cos(x + h)}{Cos(x + h)Cos(x)}\right)$

$= \lim_{h \rightarrow 0} \frac{1}{h}\left(\frac{Cos(x) - (CosxCosh - SinxSinh)}{(CosxCosh - SinxSinh)Cos(x)}\right)$

I used the Identity $Cos(x + h) = CosxCosh - SinxSinh$

After this I'm confused on where to go next. And this is supposed to be proving that $\frac{d}{dx}(Secx)=SecxTanx$ I'm able to find this with the chain rule but been a while since I used the definition :D
• Oct 14th 2009, 03:14 PM
skeeter
Quote:

Originally Posted by VitaX
Find the derivative of $Secx$ using the definition below:

$f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h}$
Note $y=Secx \rightarrow Cos^{-1}x$

$f'(Secx) = \lim_{h \rightarrow 0} \frac{Cos^{-1}(x + h) - Cos^{-1}(x)}{h}$
$f'(Secx) = \lim_{h \rightarrow 0} \frac{1}{h}\left(\frac{1}{Cos(x + h)} - \frac{1}{Cos(x)}\right)$
$f'(Secx) = \lim_{h \rightarrow 0} \frac{1}{h}\left(\frac{Cos(x) - Cos(x + h)}{Cos(x + h)Cos(x)}\right)$

After this I'm confused on where to go next. And this is supposed to be proving that $\frac{d}{dx}(Secx)=SecxTanx$ I'm able to find this with the chain rule but been a while since I used the definition :D

first off , $\cos^{-1}{x}$ is the notation for the arccosine (inverse cosine) function, not the reciprocal function $\sec{x}$.

here is the correct notation ...

$\sec{x} = (\cos{x})^{-1}$

$\lim_{h \to 0} \frac{1}{h} \left(\frac{\cos{x} - \cos(x+h)}{\cos(x+h)\cos{x}}\right)$

$\lim_{h \to 0} \frac{1}{h} \left(\frac{\cos{x} - \cos{x}\cos{h}+\sin{x}\sin{h}}{\cos(x+h)\cos{x}}\r ight)$

$\lim_{h \to 0} \frac{\cos{x}(1 - \cos{h}) + \sin{x}\sin{h}}{h \cos(x+h)\cos{x}}$

$\lim_{h \to 0} \frac{\cos{x}(1 - \cos{h})}{h \cos(x+h)\cos{x}}+ \frac{\sin{x}\sin{h}}{h \cos(x+h)\cos{x}}$

$\lim_{h \to 0} \frac{1-\cos{h}}{h} \cdot \frac{\cos{x}}{\cos(x+h)\cos{x}}+ \frac{\sin{h}}{h} \cdot \frac{\sin{x}}{\cos(x+h)\cos{x}}$

recall the two basic trig limits ...

$\lim_{t \to 0} \frac{1 - \cos{t}}{t} = 0$ and $\lim_{t \to 0} \frac{\sin{t}}{t} = 1$

$\lim_{h \to 0} \,\, 0 \cdot \frac{\cos{x}}{\cos(x+h)\cos{x}}+ 1 \cdot \frac{\sin{x}}{\cos(x+h)\cos{x}} = \frac{\sin{x}}{\cos^2{x}} = \frac{1}{\cos{x}} \cdot \frac{\sin{x}}{\cos{x}} = \sec{x} \tan{x}$
• Oct 14th 2009, 03:17 PM
VitaX
$Cos^{-1}x$ is the same as $(Cosx)^{-1}$ Atleast that is what my professor always stresses because some people get that confused with $Cosx^{-1}$ which is just the x value to that power. Thanks for the help but how come the bottom is not changed into the trig identity as well?

Quote:

Originally Posted by skeeter
first off , $\cos^{-1}{x}$ is the notation for the arccosine (inverse cosine) function, not the reciprocal function $\sec{x}$.

here is the correct notation ...

$\sec{x} = (\cos{x})^{-1}$

$\lim_{h \to 0} \frac{1}{h} \left(\frac{\cos{x} - \cos(x+h)}{\cos(x+h)\cos{x}}\right)$

$\lim_{h \to 0} \frac{1}{h} \left(\frac{\cos{x} - \cos{x}\cos{h}+\sin{x}\sin{h}}{\cos(x+h)\cos{x}}\r ight)$

$\lim_{h \to 0} \frac{\cos{x}(1 - \cos{h}) + \sin{x}\sin{h}}{h \cos(x+h)\cos{x}}$

$\lim_{h \to 0} \frac{\cos{x}(1 - \cos{h})}{h \cos(x+h)\cos{x}}+ \frac{\sin{x}\sin{h}}{h \cos(x+h)\cos{x}}$

$\lim_{h \to 0} \frac{1-\cos{h}}{h} \cdot \frac{\cos{x}}{\cos(x+h)\cos{x}}+ \frac{\sin{h}}{h} \cdot \frac{\sin{x}}{\cos(x+h)\cos{x}}$

recall the two basic trig limits ...

$\lim_{t \to 0} \frac{1 - \cos{t}}{t} = 0$ and $\lim_{t \to 0} \frac{\sin{t}}{t} = 1$

$\lim_{h \to 0} \,\, 0 \cdot \frac{\cos{x}}{\cos(x+h)\cos{x}}+ 1 \cdot \frac{\sin{x}}{\cos(x+h)\cos{x}} = \frac{\sin{x}}{\cos^2{x}} = \frac{1}{\cos{x}} \cdot \frac{\sin{x}}{\cos{x}} = \sec{x} \tan{x}$

• Oct 14th 2009, 03:19 PM
skeeter
you're welcome.
• Oct 14th 2009, 03:48 PM
VitaX
Thank you