What is the derivative of (tan^3(x)/(2x+1)) ?

The derivative of (tan^3(x)/(2x+1)) is (3tan^2(x)sec^2(x)(2x+1)-2tan^3(x))/((2x+1)^2)

What is the first derivative of (tan^3(x)/(2x+1)) ?

The first derivative of (tan^3(x)/(2x+1)) is (3tan^2(x)sec^2(x)(2x+1)-2tan^3(x))/((2x+1)^2)

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\frac{d}{d x} (\frac{tan ^{3} ( x )}{2 x + 1})

\frac{3 tan ^{2} ( x ) sec ^{2} ( x ) ( 2 x + 1 ) - 2 tan ^{3} ( x )}{( 2 x + 1 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{tan ^{3} ( x )}{2 x + 1})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d}{d x} ( tan ^{3} ( x ) ) ( 2 x + 1 ) - \frac{d}{d x} ( 2 x + 1 ) tan ^{3} ( x )}{( 2 x + 1 ) ^{2}}

\frac{d}{d x} (tan^{3} (x)) = 3 tan^{2} (x) sec^{2} (x)

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

= \frac{3 tan ^{2} ( x ) sec ^{2} ( x ) ( 2 x + 1 ) - 2 tan ^{3} ( x )}{( 2 x + 1 ) ^{2}}

\int sin (4 - x) d x

x \to 0 lim (ln ∣ sin (x) ∣)

\frac{d}{d x} (\frac{4 x ^{3}}{3} - x + 2 e^{x})

18 y^{^{''}} - 9 y^{^{'}} - 5 y = 0

\frac{d y}{d x} + 4 x y = 0

What is the derivative of (tan^3(x)/(2x+1)) ?
The derivative of (tan^3(x)/(2x+1)) is (3tan^2(x)sec^2(x)(2x+1)-2tan^3(x))/((2x+1)^2)
What is the first derivative of (tan^3(x)/(2x+1)) ?
The first derivative of (tan^3(x)/(2x+1)) is (3tan^2(x)sec^2(x)(2x+1)-2tan^3(x))/((2x+1)^2)