How do you find the derivative of #(1+x+[3x^2])^4#?

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Answer 1 · 2018-06-27T08:18:03

#4*(1+6x)(1+x+3x^2)^3#

#((f(x))^n)dx/dy=n*f'(x)*f(x)^(n-1)#

#((1+x+[3x^2])^4)dx/dy=4*(1+6x)(1+x+3x^2)^3#

Answer 2 · 2018-06-27T08:48:47

#dy/dx ->f'(x)=4(1+6x)(1+x+3x^2)^3#

Using the 'old fashioned' notation

Given: #y=(1+x+3x^2)^4#

Set #u=1+x+3x^2color(white)("d")# then #color(white)("d")(du)/dx=1+6x#

Also we have #y=u^4 color(white)("d")->color(white)("d")dy/(du)=4u^3#
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Really the format #dy/dx# is a fraction and it behaves the same way fractions do. Consequently we can multiply them.

Observe that: #(du)/dx xx dy/(du)color(white)("d") =color(white)("d") (du)/(du)xxdy/dx color(white)("d")=color(white)("d") dy/dx#

Using the above:

#dy/dx = (1+6x)xx4u^3 #

#dy/dx = (1+6x)xx4(1+x+3x^2)^3 #

#dy/dx ->f'(x)=4(1+6x)(1+x+3x^2)^3#