Calculus Derivative problem?
I'm new to calculus and derivatives and such. I can do the easy ones like:
4 - x^2
but I don't know how to do ones that involve fractions:
( 1 - X ) / ( 2 * Z )
Do I take the fraction out and find its derivative like this:
1 / ( 2 * Z) = 2z^ -1 = -2z^-2
After that I have no clue where to go withe rest of the derivative. Can someone help me please?
Update:Sorry. I was changing the variables in the problem to be X's, so all the z's in the problem SHOULD be x's.
Comments
y = (1 - x) / 2x
y = (1 / 2x) - (x / 2x)
y = (1/2)x^(-1) - 1/2
dy/dx = -(1/2)x^(-2)
dy/dx = -1 / 2x²
for (1-x)/(2x), I'd suggest rewriting it as two terms:
(1/2)x^(-1) - 1/2, ...now take derivative
[use power rule for first, constant rule for second]
(1/2)(-1)x^(-2) - 0
simplify
-1/(2x²)
OR, use the "quotient rule" if it has been introduced.
if you ask this
(1-x)/2x then you split this in 2 fractions 1/2x- x/2x so you get (1/2x)' - 2' so
(1/2x)' = 1/2 ((x)^-1 )'= 1/2(x)^-2 *(-1) = -1/2 x^(-2)
second way to do it is to use formula for fractions (here is easier to do it on first way cause you can shorten the expresion (x /2x))
second way formula
f(x)/g(x) = [f(x)'*g(x) - f(x)*g(x)']/g(x)^2
Use the product rule for products,
y=f(x)*g(x)
y'=f(x)*g'(x)+g(x)*f'(x)
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Use the quotient rule for quotients,
y'=(g(x)*f'(x)-f(x)g'(x))/(g(x))^2
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f(x)=1-x
g(x)=2x
f'(x)=-1
g'(x)=2
Using the quotient rule,
y=(1-x)/2x
y'=(2x(-1)-(1-x)2)/(4x^2)=(-2x-2+2x)/4x^2=(-2)/4x^2=-1/(2x^2)
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You could also have simplified first too,
y=(1-x)/2x=1/(2x)-1/2=(2x)^(-1)-1/2
y'=(-1)(2x)^(-2)*(2)
y'=-2/(4x^2)=-1/(2x^2)
Same answer but quicker maybe.
There are two ways to work problems with fractions.
Method 1: use the product rule or the quotient rule, and then simplify.
d/dx (f(x) * g(x)) = f(x)g'(x) + g(x)f'(x)
d/dx (1-x)(0.5x^-1) = (1-x)(-0.5x^-2) + (0.5x^ -1)(-1)
= (x-1) / (2x^2) - 1/(2x)
= -1 / (2x^2)
Method 2: simplify first and then use d/dx (f(x) + g(x)) = d/dx f(x) + d/dx g(x).
d/dx (1-x)(0.5x^-1) = d/dx (0.5x^-1 - 0.5)
= d/dx (0.5x^-1) + d/dx (-0.5)
= -0.5x^-2
= -1 / (2x^2)
(1-x)/(2x)
=(1-x)*(2x)^-1
using chain rule to derive:
[(1-x)*-1(2x)^-2*2] + [(2x)^-1*-1] <=first*d(second) + second*d(first)
simplify
-2(1-x)/4x^2 + -1/2x
get common denominator and add fractions
-2-2x+2x/4x^2
=-2/4x^2
=-1/2x^2
f = (1-x)/2x
= 1/2x - 1/2
f' = -1/2x^2