What is (\partial)/(\partial y)(-2sec^2(3x-2y+1)) ?

The answer to (\partial)/(\partial y)(-2sec^2(3x-2y+1)) is 8sec^2(3x-2y+1)tan(3x-2y+1)

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(\partial)/(\partial y)(-2sec^2(3x-2y+1))

\frac{\partial}{\partial y} (- 2 sec^{2} (3 x - 2 y + 1))

8 sec^{2} (3 x - 2 y + 1) tan (3 x - 2 y + 1)

Solution steps

\frac{\partial}{\partial y} (- 2 sec^{2} (3 x - 2 y + 1))

Treat

x

as a constant

Take the constant out:

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

= - 2 \frac{\partial}{\partial y} (sec^{2} (3 x - 2 y + 1))

Apply the chain rule:

2 sec (3 x - 2 y + 1) \frac{\partial}{\partial y} (sec (3 x - 2 y + 1))

= 2 sec (3 x - 2 y + 1) \frac{\partial}{\partial y} (sec (3 x - 2 y + 1))

\frac{\partial}{\partial y} (sec (3 x - 2 y + 1)) = - 2 sec (3 x - 2 y + 1) tan (3 x - 2 y + 1)

= - 2 \cdot 2 sec (3 x - 2 y + 1) (- 2 sec (3 x - 2 y + 1) tan (3 x - 2 y + 1))

Simplify

- 2 \cdot 2 sec (3 x - 2 y + 1) (- 2 sec (3 x - 2 y + 1) tan (3 x - 2 y + 1)) : 8 sec^{2} (3 x - 2 y + 1) tan (3 x - 2 y + 1)

= 8 sec^{2} (3 x - 2 y + 1) tan (3 x - 2 y + 1)

\frac{d}{d x} (tan (5 x + 1))

x y^{2} y^{^{'}} = x + 6

n = 1 \sum \infty \frac{7}{n ^{6}}

y = \frac{x - 1}{x + 1}

x \to - 3 lim (g (x))

What is (\partial)/(\partial y)(-2sec^2(3x-2y+1)) ?
The answer to (\partial)/(\partial y)(-2sec^2(3x-2y+1)) is 8sec^2(3x-2y+1)tan(3x-2y+1)