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The first derivative

 

        1-5x

 y = -------- =

        3x+2

Guest Feb 16, 2017
 #1
avatar+92805 
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The first derivative

 

        1-5x

 y = -------- 

        3x+2

 

\(\boxed{Quotient Rule\\ y'=\frac{vu'-uv'}{v^2}}\\~\\ y'=\frac{-5(3x+2)-3(1-5x)}{(3x+2)^2}\\ y'=\frac{-15x-10-3+5x}{(3x+2)^2}\\ y'=\frac{-10x-13}{(3x+2)^2}\\ \)

Melody  Feb 16, 2017
 #2
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Find the derivative of the following via implicit differentiation:
d/dx(y) = d/dx((1 - 5 x)/(2 + 3 x))
The derivative of y is y'(x):
y'(x) = d/dx((1 - 5 x)/(2 + 3 x))


Use the quotient rule, d/dx(u/v) = (v ( du)/( dx) - u ( dv)/( dx))/v^2, where u = 1 - 5 x and v = 3 x + 2:
y'(x) = ((3 x + 2) d/dx(1 - 5 x) - (1 - 5 x) d/dx(2 + 3 x))/(3 x + 2)^2


Differentiate the sum term by term and factor out constants:
y'(x) = (-((1 - 5 x) (d/dx(2 + 3 x))) + (2 + 3 x) d/dx(1) - 5 d/dx(x))/(2 + 3 x)^2
The derivative of 1 is zero:
y'(x) = (-((1 - 5 x) (d/dx(2 + 3 x))) + (2 + 3 x) (-5 (d/dx(x)) + 0))/(2 + 3 x)^2


Simplify the expression:
y'(x) = (-5 (2 + 3 x) (d/dx(x)) - (1 - 5 x) (d/dx(2 + 3 x)))/(2 + 3 x)^2
The derivative of x is 1:
y'(x) = (-((1 - 5 x) (d/dx(2 + 3 x))) - 1 5 (2 + 3 x))/(2 + 3 x)^2


Differentiate the sum term by term and factor out constants:
y'(x) = (-5 (2 + 3 x) - (1 - 5 x) d/dx(2) + 3 d/dx(x))/(2 + 3 x)^2
The derivative of 2 is zero:
y'(x) = (-5 (2 + 3 x) - (1 - 5 x) (3 (d/dx(x)) + 0))/(2 + 3 x)^2


Simplify the expression:
y'(x) = (-5 (2 + 3 x) - 3 (1 - 5 x) (d/dx(x)))/(2 + 3 x)^2
The derivative of x is 1:
y'(x) = (-5 (2 + 3 x) - 1 3 (1 - 5 x))/(2 + 3 x)^2


Expand the left hand side:
y'(x) = (-3 (1 - 5 x) - 5 (2 + 3 x))/(2 + 3 x)^2
Factor the numerator and denominator of the right hand side:
y'(x) = (-5 (2 + 3 x) + 3 (-1 + 5 x))/(2 + 3 x)^2
Cancel common terms in the numerator and denominator:
Answer: |y'(x) = -13/(2 + 3 x)^2

Guest Feb 16, 2017

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