What is the derivative of the function f(x) = \frac{\sin(2x)}{(1 + x^2)^2} with respect to… — JAMB UTME

What is the derivative of the function $f(x) = \frac{\sin(2x)}{(1 + x^2)^2}$ with respect to $x$ ?

\frac{2[(1 + x^2)\cos(2x) - 2x\sin(2x)]}{(1 + x^2)^3}

Applying the quotient rule

\frac{u'v - uv'}{v^2}

with

u = \sin(2x)

and

v = (1 + x^2)^2

yields

u' = 2\cos(2x)

and

v' = 4x(1 + x^2)

. Substituting these into the formula gives

\frac{2\cos(2x)(1+x^2)^2 - 4x\sin(2x)(1+x^2)}{(1+x^2)^4}

. Factoring out

(1+x^2)

from numerator and denominator reduces the expression to

\frac{2[(1 + x^2)\cos(2x) - 2x\sin(2x)]}{(1 + x^2)^3}

Identify the numerator

u(x)

and denominator

v(x)

for the quotient rule.

u(x) = \sin(2x)

and

v(x) = (1 + x^2)^2

The quotient rule states that

\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}

Find the derivatives

u'(x)

and

v'(x)

using the chain rule.

u'(x) = 2\cos(2x)

and

v'(x) = 2(1 + x^2) \cdot (2x) = 4x(1 + x^2)

By the chain rule,

\frac{d}{dx}\sin(2x) = \cos(2x) \cdot 2

and

\frac{d}{dx}(1 + x^2)^2 = 2(1 + x^2) \cdot 2x

Substitute

u, u', v, v'

into the quotient rule formula.

f'(x) = \frac{2\cos(2x)(1 + x^2)^2 - \sin(2x) \cdot 4x(1 + x^2)}{((1 + x^2)^2)^2}

Applying

\frac{u'v - uv'}{v^2}

directly.

Factor out

(1 + x^2)

from the numerator and simplify the denominator.

f'(x) = \frac{(1 + x^2)[2\cos(2x)(1 + x^2) - 4x\sin(2x)]}{(1 + x^2)^4} = \frac{2(1 + x^2)\cos(2x) - 4x\sin(2x)}{(1 + x^2)^3}

Canceling the common factor

(1 + x^2)

from numerator and denominator.

Factor out the constant

2

from the numerator.

f'(x) = \frac{2[(1 + x^2)\cos(2x) - 2x\sin(2x)]}{(1 + x^2)^3}

Putting the derivative into simplified factored form.

Combining Quotient Rule and Chain Rule for Composite Functions

Question