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

Question

2226 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-27T08:18:03

$4 \cdot (1 + 6 x) {(1 + x + 3 x^{2})}^{3}$ $4*(1+6x)(1+x+3x^2)^3$

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

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

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 $y=u^4 color(white)("d")->color(white)("d")dy/(du)=4u^3$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

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:

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

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

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

How do you find the derivative of (1+x+[3x^2])^4(1+x+[3x2])4(1+x+[3x^2])^4?