What is the derivative of y=sin(tan(9x)) ?

The derivative of y=sin(tan(9x)) is cos(tan(9x))sec^2(9x)*9

What is the first derivative of y=sin(tan(9x)) ?

The first derivative of y=sin(tan(9x)) is cos(tan(9x))sec^2(9x)*9

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derivative of

y = sin (tan (9 x))

cos (tan (9 x)) sec^{2} (9 x) \cdot 9

Solution steps

\frac{d}{d x} (sin (tan (9 x)))

Apply the chain rule:

cos (tan (9 x)) \frac{d}{d x} (tan (9 x))

= cos (tan (9 x)) \frac{d}{d x} (tan (9 x))

\frac{d}{d x} (tan (9 x)) = sec^{2} (9 x) \cdot 9

= cos (tan (9 x)) sec^{2} (9 x) \cdot 9

f (x) = - 3 x^{3} + 4, at x = - 1

\frac{\partial}{\partial x} (m x^{b})

\int_{0}^{3} x 10 - x^{2} d x

\frac{8 x}{7 - tan ( x )}

x \to \infty lim (\frac{- 1}{2})

What is the derivative of y=sin(tan(9x)) ?
The derivative of y=sin(tan(9x)) is cos(tan(9x))sec^2(9x)*9
What is the first derivative of y=sin(tan(9x)) ?
The first derivative of y=sin(tan(9x)) is cos(tan(9x))sec^2(9x)*9