What is (\partial)/(\partial x)(arcsec(x+yz)) ?

The answer to (\partial)/(\partial x)(arcsec(x+yz)) is 1/((x+yz)sqrt((x+yz)^2-1))

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(\partial)/(\partial x)(arcsec(x+yz))

\frac{\partial}{\partial x} (arcsec (x + yz))

\frac{1}{( x + yz ) ( x + yz ) ^{2} - 1}

Solution steps

\frac{\partial}{\partial x} (arcsec (x + yz))

Treat

y, z

as constants

Apply the chain rule:

\frac{1}{( x + yz ) ^{2} ( x + yz ) ^{2} - 1} \frac{\partial}{\partial x} (x + yz)

= \frac{1}{( x + yz ) ^{2} ( x + yz ) ^{2} - 1} \frac{\partial}{\partial x} (x + yz)

\frac{\partial}{\partial x} (x + yz) = 1

= \frac{1}{( x + yz ) ^{2} ( x + yz ) ^{2} - 1} \cdot 1

Simplify

\frac{1}{( x + yz ) ^{2} ( x + yz ) ^{2} - 1} \cdot 1 : \frac{1}{( x + yz ) ( x + yz ) ^{2} - 1}

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

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

\int (tan^{5} (x)) d x

f (x) = e^{x - 3}

e^{6 t}

y^{^{'}} = y - x

What is (\partial)/(\partial x)(arcsec(x+yz)) ?
The answer to (\partial)/(\partial x)(arcsec(x+yz)) is 1/((x+yz)sqrt((x+yz)^2-1))