What is the derivative of (sec(x)/(4x^2+5)) ?

The derivative of (sec(x)/(4x^2+5)) is (sec(x)tan(x)(4x^2+5)-8xsec(x))/((4x^2+5)^2)

What is the first derivative of (sec(x)/(4x^2+5)) ?

The first derivative of (sec(x)/(4x^2+5)) is (sec(x)tan(x)(4x^2+5)-8xsec(x))/((4x^2+5)^2)

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\frac{d}{d x} (\frac{sec ( x )}{4 x ^{2} + 5})

\frac{sec ( x ) tan ( x ) ( 4 x ^{2} + 5 ) - 8 x sec ( x )}{( 4 x ^{2} + 5 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{sec ( x )}{4 x ^{2} + 5})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( sec ( x ) ) ( 4 x ^{2} + 5 ) - \frac{d}{d x} ( 4 x ^{2} + 5 ) sec ( x )}{( 4 x ^{2} + 5 ) ^{2}}

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

\frac{d}{d x} (4 x^{2} + 5) = 8 x

= \frac{sec ( x ) tan ( x ) ( 4 x ^{2} + 5 ) - 8 x sec ( x )}{( 4 x ^{2} + 5 ) ^{2}}

\frac{s}{s ^{2} + 4 s + 8}

g (x) = - x^{2} + 7 x + 7

y^{^{''}} + 16 y^{^{'}} + 65 y = 0, y (0) = 1, y^{^{'}} (0) = 0

f (x) = 8 x^{\frac{1}{3}} - 5, at x = 8

3 y y^{^{'}} = x

What is the derivative of (sec(x)/(4x^2+5)) ?
The derivative of (sec(x)/(4x^2+5)) is (sec(x)tan(x)(4x^2+5)-8xsec(x))/((4x^2+5)^2)
What is the first derivative of (sec(x)/(4x^2+5)) ?
The first derivative of (sec(x)/(4x^2+5)) is (sec(x)tan(x)(4x^2+5)-8xsec(x))/((4x^2+5)^2)