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

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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Product Rule:

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

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

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

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

Simplify

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

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

ln (tan (x))

\frac{\partial}{\partial x} (\frac{( - x + y )}{x \cdot y + 3 \cdot x})

y^{^{'}} = y^{2} (5 + x)

\frac{\partial}{\partial x} (\frac{z}{1 + z ^{2} y ^{2}})

\frac{d}{d x} (A x e^{3 x})

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