What is the derivative of y=tan(ln(ax+b)) ?

The derivative of y=tan(ln(ax+b)) is (asec^2(ln(ax+b)))/(ax+b)

What is the first derivative of y=tan(ln(ax+b)) ?

The first derivative of y=tan(ln(ax+b)) is (asec^2(ln(ax+b)))/(ax+b)

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derivative of

y = tan (ln (a x + b))

\frac{a sec ^{2} ( ln ( a x + b ) )}{a x + b}

Solution steps

\frac{d}{d x} (tan (ln (a x + b)))

Apply the chain rule:

sec^{2} (ln (a x + b)) \frac{d}{d x} (ln (a x + b))

= sec^{2} (ln (a x + b)) \frac{d}{d x} (ln (a x + b))

\frac{d}{d x} (ln (a x + b)) = \frac{a}{a x + b}

= sec^{2} (ln (a x + b)) \frac{a}{a x + b}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{a sec ^{2} ( ln ( a x + b ) )}{a x + b}

y = - x + 2

(6, - 6)

x^{2} + y^{2}

f (x) = - \frac{6}{x}, at x = 12

What is the derivative of y=tan(ln(ax+b)) ?
The derivative of y=tan(ln(ax+b)) is (asec^2(ln(ax+b)))/(ax+b)
What is the first derivative of y=tan(ln(ax+b)) ?
The first derivative of y=tan(ln(ax+b)) is (asec^2(ln(ax+b)))/(ax+b)