What is (\partial)/(\partial x)(x^2tan(4x^4y)) ?

The answer to (\partial)/(\partial x)(x^2tan(4x^4y)) is 2xtan(4x^4y)+16x^5ysec^2(4x^4y)

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(\partial)/(\partial x)(x^2tan(4x^4y))

\frac{\partial}{\partial x} (x^{2} tan (4 x^{4} y))

2 x tan (4 x^{4} y) + 16 x^{5} y sec^{2} (4 x^{4} y)

Solution steps

\frac{\partial}{\partial x} (x^{2} tan (4 x^{4} y))

Treat

y

as a constant

Apply the Product Rule:

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

f = x^{2}, g = tan (4 x^{4} y)

= \frac{\partial}{\partial x} (x^{2}) tan (4 x^{4} y) + \frac{\partial}{\partial x} (tan (4 x^{4} y)) x^{2}

\frac{\partial}{\partial x} (x^{2}) = 2 x

\frac{\partial}{\partial x} (tan (4 x^{4} y)) = sec^{2} (4 x^{4} y) \cdot 16 y x^{3}

= 2 x tan (4 x^{4} y) + sec^{2} (4 x^{4} y) \cdot 16 y x^{3} x^{2}

sec^{2} (4 x^{4} y) \cdot 16 y x^{3} x^{2} = 16 x^{5} y sec^{2} (4 x^{4} y)

= 2 x tan (4 x^{4} y) + 16 x^{5} y sec^{2} (4 x^{4} y)

\frac{d}{d x} (x^{2} sin^{3} (5 x))

\int \frac{1}{x ^{2} x ^{2} - 64} d x

\int \frac{1}{t ^{2} - 8 t + 25} d t

\int e^{2 x + 5} d x

\frac{d y}{d x} = \frac{2 x}{y ^{2}}

What is (\partial)/(\partial x)(x^2tan(4x^4y)) ?
The answer to (\partial)/(\partial x)(x^2tan(4x^4y)) is 2xtan(4x^4y)+16x^5ysec^2(4x^4y)