What is derivative of f(x)=sec(θ)tan(θ) ?

The answer to derivative of f(x)=sec(θ)tan(θ) is derivative of θ=0

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derivative of f(x)=sec(θ)tan(θ)

derivative of

f (x) = sec (θ) tan (θ)

\frac{d}{d x} (θ) = 0

Solution steps

f (x) = sec (θ) tan (θ)

Treat

θ

θ (x)

Differentiate both sides:

0 = tan^{2} (θ) sec (θ) \frac{d}{d x} (θ) + sec^{3} (θ) \frac{d}{d x} (θ)

Isolate

\frac{d}{d x} (θ) : \frac{d}{d x} (θ) = 0

\frac{d}{d x} (θ) = 0

\int x^{2} (5 x^{2} + x)^{2} d x

\frac{\partial}{\partial h} (\frac{1}{2} (a + b) h)

\frac{d}{d x} (\frac{2}{x ^{\frac{1}{3}}} - 2)

f (x) = - 5 x (2 x - 3)^{4}

y^{^{''}} - 2 y^{^{'}} + 5 y = 5, y (0) = 2, y^{^{'}} (0) = 1

What is derivative of f(x)=sec(θ)tan(θ) ?
The answer to derivative of f(x)=sec(θ)tan(θ) is derivative of θ=0