Show the expanded solution on how we got the answer on this problem Product Ruleddf (x) g (x) +dg ( x) ƒ(x)dxdxdx3+ 2 x + 57Ex.2) f (x) = x3x — 8х + 1f'(x) = 3x +2 3x' -8x² +1 + x'+2x+5 21x° -16xDerivative of thefirst functionDerivative of thesecond functionf'(x)=30x +48x' +105x° –40x* –45x² –80x+2Couvrighe e 2006 Brooks Cele, a divisiou of Thomson Lexming Ine.

Given: f(x)=x3+2x+5 3x7-8x2+1 Product rule for derivative: ddxf(x). g(x)=ddxf(x) g(x)+ddxg(x)…

Product Rule d d f ( x ) g (x) + dx d · g g ( x ) ƒ(x) dx dx 3 Ex.2) f (x) = x' + 2 x + 5 3 x' - 8 x? + 1 х f'(x) = 3x +2 3x' -8x? +1 + x'+2x+5 21x –16x Derivative of the Derivative of the |first function second function f'(x)=30x +48x' +105x° –40x* –45x² –80x+2

Product Rule d d f ( x ) g (x) + dx d · g g ( x ) ƒ(x) dx dx 3 Ex.2) f (x) = x' + 2 x + 5 3 x' - 8 x? + 1 х f'(x) = 3x +2 3x' -8x? +1 + x'+2x+5 21x –16x Derivative of the Derivative of the |first function second function f'(x)=30x +48x' +105x° –40x* –45x² –80x+2

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter9: Systems Of Equations And Inequalities

Section9.10: Partial Fractions

Problem 17E

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Show the expanded solution on how we got the answer on this problem

Product Rule
d
d
f (x) g (x) +
d
g ( x) ƒ(x)
dx
dx
dx
3
+ 2 x + 5
7
Ex.2) f (x) = x
3x — 8х + 1
f'(x) = 3x +2 3x' -8x² +1 + x'+2x+5 21x° -16x
Derivative of the
first function
Derivative of the
second function
f'(x)=30x +48x' +105x° –40x* –45x² –80x+2
Couvrighe e 2006 Brooks Cele, a divisiou of Thomson Lexming Ine.

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