What is the d/(da)(sec^2(a)+tan^2(a)) ?

The d/(da)(sec^2(a)+tan^2(a)) is 4sec^2(a)tan(a)

What is the first d/(da)(sec^2(a)+tan^2(a)) ?

The first d/(da)(sec^2(a)+tan^2(a)) is 4sec^2(a)tan(a)

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

4 sec^{2} (a) tan (a)

Solution steps

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

Simplify

sec^{2} (a) + tan^{2} (a) : 1 + 2 tan^{2} (a)

= \frac{d}{d a} (1 + 2 tan^{2} (a))

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d a} (1) + \frac{d}{d a} (2 tan^{2} (a))

\frac{d}{d a} (1) = 0

\frac{d}{d a} (2 tan^{2} (a)) = 4 sec^{2} (a) tan (a)

= 0 + 4 sec^{2} (a) tan (a)

Simplify

= 4 sec^{2} (a) tan (a)

y = 5 sin (x) + 3 cos (x) + 5 x^{2}

\frac{d}{d n} (i^{4 n + 1})

ln ((2 x + 3)^{2})

\frac{1}{( ( 1 + x ) ^{2} ln ^{3} ( x + 1 ) )}

\frac{2 x - 4}{x ^{\frac{1}{2}}}

What is the d/(da)(sec^2(a)+tan^2(a)) ?
The d/(da)(sec^2(a)+tan^2(a)) is 4sec^2(a)tan(a)
What is the first d/(da)(sec^2(a)+tan^2(a)) ?
The first d/(da)(sec^2(a)+tan^2(a)) is 4sec^2(a)tan(a)