What is the derivative of 4sin(tan(2x+1)) ?

The derivative of 4sin(tan(2x+1)) is 8sec^2(2x+1)cos(tan(2x+1))

What is the first derivative of 4sin(tan(2x+1)) ?

The first derivative of 4sin(tan(2x+1)) is 8sec^2(2x+1)cos(tan(2x+1))

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\frac{d}{d x} (4 sin (tan (2 x + 1)))

8 sec^{2} (2 x + 1) cos (tan (2 x + 1))

Solution steps

\frac{d}{d x} (4 sin (tan (2 x + 1)))

Take the constant out:

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

= 4 \frac{d}{d x} (sin (tan (2 x + 1)))

Apply the chain rule:

cos (tan (2 x + 1)) \frac{d}{d x} (tan (2 x + 1))

= cos (tan (2 x + 1)) \frac{d}{d x} (tan (2 x + 1))

\frac{d}{d x} (tan (2 x + 1)) = sec^{2} (2 x + 1) \cdot 2

= 4 cos (tan (2 x + 1)) sec^{2} (2 x + 1) \cdot 2

Simplify

= 8 sec^{2} (2 x + 1) cos (tan (2 x + 1))

\int x \cdot (x + 5)^{4} d x

\int (3 x^{2} + 1)^{3} 6 x d x

\int \frac{a ^{2}}{x x ^{2} - a ^{2}} d x

\int (3 x - 2) e^{3 x} d x

a re a x^{6}, [0, 1]

What is the derivative of 4sin(tan(2x+1)) ?
The derivative of 4sin(tan(2x+1)) is 8sec^2(2x+1)cos(tan(2x+1))
What is the first derivative of 4sin(tan(2x+1)) ?
The first derivative of 4sin(tan(2x+1)) is 8sec^2(2x+1)cos(tan(2x+1))