Question

Find dy: y =

dy
=
x3 +52 +4
x²+4
Calculator
Check Answer
Jump to Answer

Answer 1

Given:

`y=(x^3+5x+4)/(x^2+4)`

Find:

`dy=??`

Rules/Methods that will be used:

Product rule:
`\Longrightarrow (d)/(dx)[(f(x))/(g(x)) ] = (g(x)f'(x)-f(x)g'(x))/((g(x))^2)`

Power Rule:
`\Longrightarrow (d)/(dx)[x^n]=nx^(n-1)`

Constant Rule:
`\Longrightarrow (d)/(dx)[k]=0`

`\Longrightarrow y=(x^3+5x+4)/(x^2+4) \Longrightarrow dy=((x^2+4)((d)/(dx)[x^3+5x+4])-(x^3+5x+4)((d)/(dx)[x^2+4] ))/((x^2+4)^2)`

`\Longrightarrow dy=((x^2+4)(3x^2+5)-(x^3+5x+4)(2x))/((x^2+4)^2)`

Simplify the expression.

`\Longrightarrow dy=((x^2+4)(3x^2+5)-(x^3+5x+4)(2x))/((x^2+4)^2) \Longrightarrow dy=((3x^4+17x^2+20)-(2x^4+10x^2+8x))/((x^2+4)^2)`

`\Longrightarrow dy=(x^4+7x^2-8x+20)/((x^2+4)^2)`

The expression cannot be simplified further.

`Thus, dy=\boxed{(x^4+7x^2-8x+20)/((x^2+4)^2)} \therefore Sol.`