What is the derivative of sin((5x-3x^3/(x+1))) ?

The derivative of sin((5x-3x^3/(x+1))) is (cos(\frac{5x-3x^3)/(x+1))(-6x^3-9x^2+5)}{(x+1)^2}

What is the first derivative of sin((5x-3x^3/(x+1))) ?

The first derivative of sin((5x-3x^3/(x+1))) is (cos(\frac{5x-3x^3)/(x+1))(-6x^3-9x^2+5)}{(x+1)^2}

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\frac{d}{d x} (sin (\frac{5 x - 3 x ^{3}}{x + 1}))

\frac{cos ( \frac{5 x - 3 x ^{3}}{x + 1} ) ( - 6 x ^{3} - 9 x ^{2} + 5 )}{( x + 1 ) ^{2}}

Solution steps

\frac{d}{d x} (sin (\frac{5 x - 3 x ^{3}}{x + 1}))

Apply the chain rule:

cos (\frac{5 x - 3 x ^{3}}{x + 1}) \frac{d}{d x} (\frac{5 x - 3 x ^{3}}{x + 1})

= cos (\frac{5 x - 3 x ^{3}}{x + 1}) \frac{d}{d x} (\frac{5 x - 3 x ^{3}}{x + 1})

\frac{d}{d x} (\frac{5 x - 3 x ^{3}}{x + 1}) = \frac{- 6 x ^{3} - 9 x ^{2} + 5}{( x + 1 ) ^{2}}

= cos (\frac{5 x - 3 x ^{3}}{x + 1}) \frac{- 6 x ^{3} - 9 x ^{2} + 5}{( x + 1 ) ^{2}}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{cos ( \frac{5 x - 3 x ^{3}}{x + 1} ) ( - 6 x ^{3} - 9 x ^{2} + 5 )}{( x + 1 ) ^{2}}

x^{n}

\int \frac{3 x ^{2} + x + 1}{x ^{2} - 2 x + 5} d x

arctan (\frac{2}{x})

x^{2} y^{^{''}} + y^{^{'}} = 0

\int_{0}^{1} x^{6} (1 + 2 x^{7})^{5} d x

What is the derivative of sin((5x-3x^3/(x+1))) ?
The derivative of sin((5x-3x^3/(x+1))) is (cos(\frac{5x-3x^3)/(x+1))(-6x^3-9x^2+5)}{(x+1)^2}
What is the first derivative of sin((5x-3x^3/(x+1))) ?
The first derivative of sin((5x-3x^3/(x+1))) is (cos(\frac{5x-3x^3)/(x+1))(-6x^3-9x^2+5)}{(x+1)^2}