What is (\partial)/(\partial x)(sqrt(x^2-ln(xy))) ?

The answer to (\partial)/(\partial x)(sqrt(x^2-ln(xy))) is (2x^2-1)/(2xsqrt(x^2-ln(xy)))

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(\partial)/(\partial x)(sqrt(x^2-ln(xy)))

\frac{\partial}{\partial x} (x^{2} - ln (x y))

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

Solution steps

\frac{\partial}{\partial x} (x^{2} - ln (x y))

Treat

y

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d}{d x} (π csc (π + \frac{x ^{6}}{3}))

f (x) = (x - 1)

y^{^{''}} + 4 y^{^{'}} + 4 y = \frac{( e ^{- 2 x} )}{( x ^{2} )}

\int 5 x^{4} - sin (x) + 2 e^{x} d x

\int \frac{x ^{4}}{3 - x ^{2}} d x

What is (\partial)/(\partial x)(sqrt(x^2-ln(xy))) ?
The answer to (\partial)/(\partial x)(sqrt(x^2-ln(xy))) is (2x^2-1)/(2xsqrt(x^2-ln(xy)))