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

The derivative of (tan(x)/(x^3)) is (xsec^2(x)-3tan(x))/(x^4)

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

The first derivative of (tan(x)/(x^3)) is (xsec^2(x)-3tan(x))/(x^4)

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\frac{d}{d x} (\frac{tan ( x )}{x ^{3}})

\frac{x sec ^{2} ( x ) - 3 tan ( x )}{x ^{4}}

Solution steps

\frac{d}{d x} (\frac{tan ( x )}{x ^{3}})

Apply the Quotient Rule:

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

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

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

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

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

Simplify

\frac{sec ^{2} ( x ) x ^{3} - 3 x ^{2} tan ( x )}{( x ^{3} ) ^{2}} : \frac{x sec ^{2} ( x ) - 3 tan ( x )}{x ^{4}}

= \frac{x sec ^{2} ( x ) - 3 tan ( x )}{x ^{4}}

(f (5 x))^{2} - 2

\int \frac{sec ^{2} ( \frac{1}{x ^{5}} )}{x ^{6}} d x

\int \frac{1}{( 16 + x ^{2} ) ^{\frac{3}{2}}} d x

a x + b

\int_{1}^{4} x^{2} - 3 x + 12 d x

What is the derivative of (tan(x)/(x^3)) ?
The derivative of (tan(x)/(x^3)) is (xsec^2(x)-3tan(x))/(x^4)
What is the first derivative of (tan(x)/(x^3)) ?
The first derivative of (tan(x)/(x^3)) is (xsec^2(x)-3tan(x))/(x^4)