What is (\partial)/(\partial x)(2sec^2(2x-y)) ?

The answer to (\partial)/(\partial x)(2sec^2(2x-y)) is 8sec^2(2x-y)tan(2x-y)

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(\partial)/(\partial x)(2sec^2(2x-y))

\frac{\partial}{\partial x} (2 sec^{2} (2 x - y))

8 sec^{2} (2 x - y) tan (2 x - y)

Solution steps

\frac{\partial}{\partial x} (2 sec^{2} (2 x - y))

Treat

y

as a constant

Take the constant out:

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

= 2 \frac{\partial}{\partial x} (sec^{2} (2 x - y))

Apply the chain rule:

2 sec (2 x - y) \frac{\partial}{\partial x} (sec (2 x - y))

= 2 sec (2 x - y) \frac{\partial}{\partial x} (sec (2 x - y))

\frac{\partial}{\partial x} (sec (2 x - y)) = sec (2 x - y) tan (2 x - y) \cdot 2

= 2 \cdot 2 sec (2 x - y) sec (2 x - y) tan (2 x - y) \cdot 2

Simplify

2 \cdot 2 sec (2 x - y) sec (2 x - y) tan (2 x - y) \cdot 2 : 8 sec^{2} (2 x - y) tan (2 x - y)

= 8 sec^{2} (2 x - y) tan (2 x - y)

2 x + y^{2} + 2 x y y^{^{'}} = 0

\frac{d}{d x} (x - ln (x + 1))

\int cos^{7} (9 - x) sin (9 - x) d x

v^{'} - 2 = 2 v^{2}

\frac{d}{d x} ((2 x + 7)^{3})

What is (\partial)/(\partial x)(2sec^2(2x-y)) ?
The answer to (\partial)/(\partial x)(2sec^2(2x-y)) is 8sec^2(2x-y)tan(2x-y)