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

The answer to (\partial)/(\partial x)(3sec^2(3x-2y+1)) is 18sec^2(3x-2y+1)tan(3x-2y+1)

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

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

Simplify

3 \cdot 2 sec (3 x - 2 y + 1) sec (3 x - 2 y + 1) tan (3 x - 2 y + 1) \cdot 3 : 18 sec^{2} (3 x - 2 y + 1) tan (3 x - 2 y + 1)

= 18 sec^{2} (3 x - 2 y + 1) tan (3 x - 2 y + 1)

\frac{d}{d x} ((2 x + 3)^{5} + 10 x (2 x + 3)^{4})

21 x y^{^{'}} + y = 462 x

f (x) = \frac{1}{x + 3}, at x = 2

\int x^{3} (7 + x^{4})^{5} d x

\int_{0}^{\frac{π}{3}} ln (sec (x)) d x

What is (\partial)/(\partial x)(3sec^2(3x-2y+1)) ?
The answer to (\partial)/(\partial x)(3sec^2(3x-2y+1)) is 18sec^2(3x-2y+1)tan(3x-2y+1)