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

The derivative of tan(ax+b) is sec^2(ax+b)a

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

The first derivative of tan(ax+b) is sec^2(ax+b)a

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

sec^{2} (a x + b) a

Solution steps

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

Apply the chain rule:

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

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

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

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

\frac{\partial}{\partial x} (3 y^{2} x - y^{3} - x^{4})

x^{2} y^{^{''}} = y^{^{'}} + (y^{^{'}})^{2}

\frac{\partial}{\partial x} (x^{x y})

\int \frac{1}{x ^{4} - 16} d x

\int_{0}^{1} 5 e^{5 x} d x

What is the derivative of tan(ax+b) ?
The derivative of tan(ax+b) is sec^2(ax+b)a
What is the first derivative of tan(ax+b) ?
The first derivative of tan(ax+b) is sec^2(ax+b)a