derivative/chainrule

Question

y=(4x-1)^5(2x+3)^4 find y' can someone explain how to do this in steps please?

Answer 1

Using the product rule:

dy/dx = [d/dx ((4x-1)^5)]*[(2x+3)^4] +[(4x-1)^5]*[d/dx ((2x+3)^4)]

Using power and chain rules twice:

dy/dx = (5(4x-1)^4)*[d/dx (4x-1)]*[(2x+3)^4]

             + [(4x-1)^5]*(4(2x+3)^3)*[d/dx (2x+3)]

Applying linear differentiation twice:

dy/dx = (5(4x-1)^4)*[4]*[(2x+3)^4] + [(4x-1)^5]*(4(2x+3)^3)*[2]

Algebraic simplification:

dy/dx = 20*((4x-1)^4)*((2x+3)^4) + 8*((4x-1)^5)*((2x+3)^3)

Greatest common factor:

dy/dx = 4*((4x-1)^4)*((2x+3)^3)*(5*(2x+3) + 2*(4x-1))

Algebraic simplification:
dy/dx = 4*((4x-1)^4)*((2x+3)^3)*( 10x+15 + 8x-2)
dy/dx = 4*((4x-1)^4)*((2x+3)^3)*( 18x+13 )

 

Leaving the derivative in factored form makes it easier to find and graph the zeros of this 8th degree polynomial; then to sketch its graph (it's even with both end behaviors up).

Answer 2

You have both product rule and chain rule here.

 

y'(x)=[(4x-1)⁵(2x+3)⁴]'=[(4x-1)⁵]'*(2x+3)⁴+(4x-1)⁵*[(2x+3)⁴]'

 

This is product rule application. Now let me work out [(4x-1)⁵]'

 

Recall chain rule:

 

f(y(x))'=f'(y(x))_y*y'(x)_x=df/dy*dy/dx

 

In our case, y(x)=4x-1; f(y)=y⁵;

df/dy=f'(y(x))=(y⁵)'=5y⁴=5(4x-1)⁴

dy/dx=y'(x)=(4x-1)'=4.

Thus, f'(x)=5(4x-1)⁴*4=20(4x-1)⁴

 

Analogously, for [(2x+3)⁴]':

[(2x+3)⁴]'=4(2x+3)³*2=8(2x+3)³

 

Total derivative is:

[(4x-1)⁵(2x+3)⁴]'=20(4x-1)⁴(2x+3)⁴+8(4x-1)⁵(2x+3)³=4(4x-1)⁴(2x+3)³(10x+15+8x-2)=

=4(4x-1)⁴(2x+3)³(18x+13)

 

Answer: 4(4x-1)⁴(2x+3)³(18x+13)