What is the (2sec^2(x)tan(x))^' ?

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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

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

\int \frac{1}{( x + 2 ) x ^{2} + 4 x - 12} d x

\frac{1}{1 + x ^{2}}

x \to 0 - lim (6 x - 1)

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

x \to 0 lim (\frac{x + ∣ x ∣}{x})

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