definition of the derivative

[Sophie]

I have to find the slope of the tangent line to the function

y=1/(Sqrt(2x-1))

at the point a using the definition of the derivative.

I am confused whether the definition of the derivative is

m=lim x=a (f(x)-f(a))/(x-a)

or

m=lim h=0 (f(a=h) - f(a))/h

or both.

I could not find the term derivatice in the text I am using.

Thanks sophie

 

[galactus]

Definition of derivative:

\(\displaystyle \L\\\lim_{h\to\0}\frac{f(x+h)-f(x)}{h}\)

You have:

\(\displaystyle \L\\\lim_{h\to\0}\frac{\frac{1}{\sqrt{2(x+h)-1}}-\frac{1}{\sqrt{2x-1}}}{h}\)

Multiply the top and bottom by the conjugate of the numerator:

\(\displaystyle \L\\\lim_{h\to\0}\frac{\frac{1}{\sqrt{2(x+h)-1}}-\frac{1}{\sqrt{2x-1}}}{h}\cdot\frac{\frac{1}{\sqrt{2(x+h)-1}}+\frac{1}{\sqrt{2x-1}}}{\frac{1}{\sqrt{2(x+h)-1}}+\frac{1}{\sqrt{2x-1}}\)

This gives:

\(\displaystyle \L\\\lim_{h\to\0}\frac{-2}{\sqrt{2x-1}\sqrt{2x+2h-1}(\sqrt{2x+2h-1}+\sqrt{2x-1})}\)

As \(\displaystyle {h\to\0}\):

\(\displaystyle \L\\\frac{-2}{\sqrt{2x-1}\sqrt{2x+\sout{2h}-1}(\sqrt{2x+\sout{2h}-1}+\sqrt{2x-1})}\)

\(\displaystyle \L\\=\frac{-2}{(2x-1)(2\sqrt{2x-1})}\)

\(\displaystyle \H\\=\frac{-1}{(2x-1)^{\frac{3}{2}}}\)


You can do the conjugate thing with the other form too.

 

[arthur ohlsten]

derivative of f[x]= limit as h approaches 0 of { f[x+h]-f[x] } /h
y= 1 / sqrt[2x-1]
[y+h]= 1/sqrt[2[x+h]-1] subtract first from second equation
[y+h]-y= 1/sqrt[2[x+h]-1] - 1/sqrt[2x-1]
[y+h]-y= {sqrt [2x-1] - sqrt[2x-1]+2h] } /sqrt[2x+2h-1][2x-1]
by binomial expansion of sqrt[[2x-1]+2h]
[y+h]-y]= sqrt[2x-1] - {[2x-1]^1/2 +1/2 [2x-1]^-1/2[2h] + higher orders of h}
divided by sqrt[2x+2h-1][2x-1]

divide by h
{[y+h]-y}/h= [2x-1]^1/2 /h - [2x-1]^1/2 /h - 1/2 [2x-1]^-1/2[2h]/h +highorders
divided by sqrt[2x+2h-1][2x-1]

{[y+h]-y}/h= -1/2 [2x-1]^-1/2 [2]+ higher orders of h / sqrt[[2x-1]^2+2h[2x-1]]
limit as h-->0
dy/dx= -1[2x-1]^1/2 / sqrt[2x-1]^2
dy/dx = - 1/[2x-1]3/2 answer

Arthur