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

The answer to (\partial)/(\partial x)(sec^2(2x+3y)*3) is 12sec^2(2x+3y)tan(2x+3y)

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

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

12 sec^{2} (2 x + 3 y) tan (2 x + 3 y)

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

Simplify

3 \cdot 2 sec (2 x + 3 y) sec (2 x + 3 y) tan (2 x + 3 y) \cdot 2 : 12 sec^{2} (2 x + 3 y) tan (2 x + 3 y)

= 12 sec^{2} (2 x + 3 y) tan (2 x + 3 y)

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

\frac{d}{d x} ((\frac{x ^{3} - 1}{x ^{3} + 1})^{8})

y = - 3 x^{2} + 2

x \to 0 lim (\frac{5 x}{tan ( x )})

\int 2 + tan^{2} (θ) d θ

What is (\partial)/(\partial x)(sec^2(2x+3y)*3) ?
The answer to (\partial)/(\partial x)(sec^2(2x+3y)*3) is 12sec^2(2x+3y)tan(2x+3y)