Answer to Question #305571 in Calculus for Anniee

Question

Use the rules of differentiation to differentiate the following functions.

a. f(x)=2x²+6x

b.g(x)=7x⁴-3x²

c.y(x)=(4x)³- 18x²+6x

d.h(x)=(3x+4)²

e.h(x)=9x⅔+2/4√x

Answer 1

a. $f(x)=2x^2+6x$

here we apply the sum rule

$d/dx(f(x)) =d/dx(g(x))+d/dx(h(x))$

let $g(x)= 2x^2$

$h(x)=6x$

$=d/dx(2x^3)+d/dx(6x)$

$=6x^2+6$

b. g(x)= $7x^4-3x^2$

here we apply the difference rule

$d/dx(g(x)) =d/dx(f(x))-d/dx(h(x))$

let $f(x)= 7x^4$

$h(x)=3x^2$

$=d/dx(7x^4)-d/dx(3x^2)$

$=28x^3-6x$

c. $y(x)=(4x)^3-18x^2+6x$

we start by simplifying the equation

= $(4x)^3=64x^3$

the simplified equation will be

$64x^3-18x^2+6x$

here we apply the sum/ difference rule

$$d/dx(y(x)) =d/dx(f(x)) -d/dx(g(x))+d/dx(h(x))$$

let $f(x)= 64x^3$

$g(x) = 18x^2$

$h(x)=6x$

$=d/dx(y(x))=d/dx(64x^3)-d/dx(18x^2)+d/dx(6x)$

$=192x^2-36x+6$

d. $h(x)=(3x+4)^2$

we apply the chain rule

$d/dx[f(g(x)] =d/d[g(x)][f(x)]*d/dx(g(x))$

let f(x)= 2

$g(x)= 3x+4$

=$2*(3x+4)*d/dx(3x+4)$

= $2*(3x+4)*3$

=$6(3x+4)$

= $18x+24$

e. $h(x)=9x$^2/3+$2/4 ​ \sqrt{x}$

here we apply the sum rule.

d/dx(h(x))=d/dx(g(x))+d/dx(h(x))

=$d/dx($ 9x^2/3)+$d/dx$ ($2/4 ​ \sqrt{x}$)

= 6x ^-1/3-1/4x^3/2