What is the derivative of 2sec^2(x)tan(x) ?

The derivative of 2sec^2(x)tan(x) is -4sec^2(x)+6sec^4(x)

What is the first derivative of 2sec^2(x)tan(x) ?

The first derivative of 2sec^2(x)tan(x) is -4sec^2(x)+6sec^4(x)

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derivative of

2 sec^{2} (x) tan (x)

- 4 sec^{2} (x) + 6 sec^{4} (x)

Solution steps

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

Take the constant out:

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

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = sec^{2} (x), g = tan (x)

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

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

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

= 2 (2 sec^{2} (x) tan (x) tan (x) + sec^{2} (x) sec^{2} (x))

Simplify

2 (2 sec^{2} (x) tan (x) tan (x) + sec^{2} (x) sec^{2} (x)) : - 4 sec^{2} (x) + 6 sec^{4} (x)

= - 4 sec^{2} (x) + 6 sec^{4} (x)

cot (x^{2} + π x)

\frac{s + 1}{( s ^{2} + 2 ) ( s + 1 )}

9 sin (2 x)

f (x) = \frac{x}{4 + x ^{2}}

(4^{x})^{^{'}}

What is the derivative of 2sec^2(x)tan(x) ?
The derivative of 2sec^2(x)tan(x) is -4sec^2(x)+6sec^4(x)
What is the first derivative of 2sec^2(x)tan(x) ?
The first derivative of 2sec^2(x)tan(x) is -4sec^2(x)+6sec^4(x)