What is the derivative of arcsec(5ln(2x)) ?

The derivative of arcsec(5ln(2x)) is 1/(xsqrt(ln^2(2x))\sqrt{25ln^2(2x)-1)}

What is the first derivative of arcsec(5ln(2x)) ?

The first derivative of arcsec(5ln(2x)) is 1/(xsqrt(ln^2(2x))\sqrt{25ln^2(2x)-1)}

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derivative of

arcsec (5 ln (2 x))

\frac{1}{x ln ^{2} ( 2 x ) 25 ln ^{2} ( 2 x ) - 1}

Solution steps

\frac{d}{d x} (arcsec (5 ln (2 x)))

Apply the chain rule:

\frac{1}{( 5 ln ( 2 x ) ) ^{2} ( 5 ln ( 2 x ) ) ^{2} - 1} \frac{d}{d x} (5 ln (2 x))

= \frac{1}{( 5 ln ( 2 x ) ) ^{2} ( 5 ln ( 2 x ) ) ^{2} - 1} \frac{d}{d x} (5 ln (2 x))

\frac{d}{d x} (5 ln (2 x)) = \frac{5}{x}

= \frac{1}{( 5 ln ( 2 x ) ) ^{2} ( 5 ln ( 2 x ) ) ^{2} - 1} \cdot \frac{5}{x}

Simplify

\frac{1}{( 5 ln ( 2 x ) ) ^{2} ( 5 ln ( 2 x ) ) ^{2} - 1} \cdot \frac{5}{x} : \frac{1}{x ln ^{2} ( 2 x ) 25 ln ^{2} ( 2 x ) - 1}

= \frac{1}{x ln ^{2} ( 2 x ) 25 ln ^{2} ( 2 x ) - 1}

\frac{d x}{d y}, y^{2} = x

\int \frac{1}{64 + ( x - 1 ) ^{2}} d x

\frac{d y}{d x} = 50 y - \frac{1}{10} y^{2}

x^{2} + 1

\int (\frac{1}{5 x ^{3.3}}) d x

What is the derivative of arcsec(5ln(2x)) ?
The derivative of arcsec(5ln(2x)) is 1/(xsqrt(ln^2(2x))\sqrt{25ln^2(2x)-1)}
What is the first derivative of arcsec(5ln(2x)) ?
The first derivative of arcsec(5ln(2x)) is 1/(xsqrt(ln^2(2x))\sqrt{25ln^2(2x)-1)}