What is (\partial)/(\partial x)(sqrt(x^y+ln(z))) ?

The answer to (\partial)/(\partial x)(sqrt(x^y+ln(z))) is (yx^{y-1})/(2sqrt(x^y+ln(z)))

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(\partial)/(\partial x)(sqrt(x^y+ln(z)))

\frac{\partial}{\partial x} (x^{y} + ln (z))

\frac{y x ^{y - 1}}{2 x ^{y} + ln ( z )}

Solution steps

\frac{\partial}{\partial x} (x^{y} + ln (z))

Treat

y, z

as constants

Apply the chain rule:

\frac{1}{2 x ^{y} + ln ( z )} \frac{\partial}{\partial x} (x^{y} + ln (z))

= \frac{1}{2 x ^{y} + ln ( z )} \frac{\partial}{\partial x} (x^{y} + ln (z))

\frac{\partial}{\partial x} (x^{y} + ln (z)) = y x^{y - 1}

= \frac{1}{2 x ^{y} + ln ( z )} y x^{y - 1}

Simplify

\frac{1}{2 x ^{y} + ln ( z )} y x^{y - 1} : \frac{y x ^{y - 1}}{2 x ^{y} + ln ( z )}

= \frac{y x ^{y - 1}}{2 x ^{y} + ln ( z )}

sec (x) \frac{d y}{d x} = e^{y + s i n (x)}

\int \frac{x}{x ^{2} + 7} d x

\frac{\partial}{\partial x} (- 4 x^{2})

(- e^{x})^{^{'}}

2 x^{2} y^{^{'}} = y^{^{'}} + 6 x e^{- y}

What is (\partial)/(\partial x)(sqrt(x^y+ln(z))) ?
The answer to (\partial)/(\partial x)(sqrt(x^y+ln(z))) is (yx^{y-1})/(2sqrt(x^y+ln(z)))