What is the derivative of-tan(2sin(4x)) ?

The derivative of-tan(2sin(4x)) is -sec^2(2sin(4x))*8cos(4x)

What is the first derivative of-tan(2sin(4x)) ?

The first derivative of-tan(2sin(4x)) is -sec^2(2sin(4x))*8cos(4x)

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derivative of

- tan (2 sin (4 x))

- sec^{2} (2 sin (4 x)) \cdot 8 cos (4 x)

Solution steps

\frac{d}{d x} (- tan (2 sin (4 x)))

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= - \frac{d}{d x} (tan (2 sin (4 x)))

Apply the chain rule:

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

= sec^{2} (2 sin (4 x)) \frac{d}{d x} (2 sin (4 x))

\frac{d}{d x} (2 sin (4 x)) = 8 cos (4 x)

= - sec^{2} (2 sin (4 x)) \cdot 8 cos (4 x)

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

x \to 2 lim (\frac{2}{x ^{2} - 4})

\int \frac{- 2 x}{3} d x

y^{^{''}} - 9 y^{^{'}} = 1 + 3 e^{2 x}

\frac{d y}{d x} = \frac{y}{x + 1}

What is the derivative of-tan(2sin(4x)) ?
The derivative of-tan(2sin(4x)) is -sec^2(2sin(4x))*8cos(4x)
What is the first derivative of-tan(2sin(4x)) ?
The first derivative of-tan(2sin(4x)) is -sec^2(2sin(4x))*8cos(4x)