What is (\partial)/(\partial x)(sec^5(x^3y^3)) ?

The answer to (\partial)/(\partial x)(sec^5(x^3y^3)) is 15x^2y^3sec^5(x^3y^3)tan(x^3y^3)

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(\partial)/(\partial x)(sec^5(x^3y^3))

\frac{\partial}{\partial x} (sec^{5} (x^{3} y^{3}))

15 x^{2} y^{3} sec^{5} (x^{3} y^{3}) tan (x^{3} y^{3})

Solution steps

\frac{\partial}{\partial x} (sec^{5} (x^{3} y^{3}))

Treat

y

as a constant

Apply the chain rule:

5 (sec (x^{3} y^{3}))^{4} \frac{\partial}{\partial x} (sec (x^{3} y^{3}))

= 5 (sec (x^{3} y^{3}))^{4} \frac{\partial}{\partial x} (sec (x^{3} y^{3}))

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

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

Simplify

5 sec^{4} (x^{3} y^{3}) sec (x^{3} y^{3}) tan (x^{3} y^{3}) \cdot 3 y^{3} x^{2} : 15 x^{2} y^{3} sec^{5} (x^{3} y^{3}) tan (x^{3} y^{3})

= 15 x^{2} y^{3} sec^{5} (x^{3} y^{3}) tan (x^{3} y^{3})

a sin (x) + b cos (x)

f (x) = x + 1 + \frac{1}{x ^{2}}

\frac{d}{d x} (tan^{3} (2 x))

\int \frac{7}{2 x ^{\frac{9}{4}}} d x

y^{^{''}} - 8 y^{^{'}} + 16 y = t^{- 9} e^{4 t}

What is (\partial)/(\partial x)(sec^5(x^3y^3)) ?
The answer to (\partial)/(\partial x)(sec^5(x^3y^3)) is 15x^2y^3sec^5(x^3y^3)tan(x^3y^3)