Differentiation question?

Case I : 1/(x² + 2)

(d/dx)[1/(x² + 2)]
= d(x² + 2)⁻¹/dx
= [d(x² + 2)⁻¹/d(x² + 2)] * [d(x² + 2)/dx]
= [-(x² + 2)⁻²] * (2x)
= -2x/(x² + 2)²


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Case II : (1/x²) + 2

(d/dx)[ (1/x²) + 2]
= [(d x⁻²)/dx] + (d2/dx)
= -2x⁻³
= -2/x³

Hello,

y = 1 /(x² + 2)

let's rewrite it as:

y = (x² + 2)‾ ¹

let's differentiate applying the power rule (xⁿ)’ = n xⁿ ‾ ¹ and the chain rule (that is also multiplying by the derivative of (x² + 2) ):

y' = (- 1)(x² + 2)‾ ¹ ‾ ¹ (x² + 2)' =

(applying the power rule (xⁿ)’ = n xⁿ ‾ ¹)

- (x² + 2)‾ ² (2x² ‾ ¹ + 0) =

- [1 /(x² + 2)²] 2x =

ending with:


- 2x /(x² + 2)²



(if instead you meant:

y = (1/x²) + 2

let's rewrite it as:

y = x‾ ² + 2

then let's simply apply the power rule (xⁿ)’ = n xⁿ ‾ ¹:

y' = - 2x‾ ² ‾ ¹ + 0 =

- 2x‾ ³ =


- 2/x³ )



I hope it's helpful

Assuming you mean 1/(x² + 2)..we can use the quotient rule

i.e. f(x) = u/v, where f '(x) = (u'v - uv')/v²

Hence, if u = 1, u' = 0, v = x² + 2, v' = 2x

so, f '(x) => (-2x)/(x² + 2)²

:)>

As given MUST be read as -

f (x) = [ 1/x² ] + 2
f ` (x) = - 2/x³

However it may be that what you ACTUALLY mean is -

f (x) = 1 / [ x³ + 2 ] = [ x³ + 2 ]^(-1)
f ` (x) = - [ x³ + 2 ]^(-2) (3x²)
f ` (x) = (- 3x²) / [ x³ + 2 ]²