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

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

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

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

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

\frac{d}{d x} (x^{2}) = 2 x

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

Simplify

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

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

0.0018425 (t + 7)^{4.5}

y = x^{2} ln (2 x)

\int \frac{1 - x}{x} d x

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

\frac{d}{d x} (2 x - y)

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