What is (\partial)/(\partial x)(ln(1+(x-1)^4*y^2)) ?

The answer to (\partial)/(\partial x)(ln(1+(x-1)^4*y^2)) is (4y^2(x-1)^3)/(1+y^2(x-1)^4)

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(\partial)/(\partial x)(ln(1+(x-1)^4*y^2))

\frac{\partial}{\partial x} (ln (1 + (x - 1)^{4} \cdot y^{2}))

\frac{4 y ^{2} ( x - 1 ) ^{3}}{1 + y ^{2} ( x - 1 ) ^{4}}

Solution steps

\frac{\partial}{\partial x} (ln (1 + (x - 1)^{4} y^{2}))

Treat

y

as a constant

Apply the chain rule:

\frac{1}{1 + ( x - 1 ) ^{4} y ^{2}} \frac{\partial}{\partial x} (1 + (x - 1)^{4} y^{2})

= \frac{1}{1 + ( x - 1 ) ^{4} y ^{2}} \frac{\partial}{\partial x} (1 + (x - 1)^{4} y^{2})

\frac{\partial}{\partial x} (1 + (x - 1)^{4} y^{2}) = 4 y^{2} (x - 1)^{3}

= \frac{1}{1 + ( x - 1 ) ^{4} y ^{2}} \cdot 4 y^{2} (x - 1)^{3}

Simplify

\frac{1}{1 + ( x - 1 ) ^{4} y ^{2}} \cdot 4 y^{2} (x - 1)^{3} : \frac{4 y ^{2} ( x - 1 ) ^{3}}{1 + y ^{2} ( x - 1 ) ^{4}}

= \frac{4 y ^{2} ( x - 1 ) ^{3}}{1 + y ^{2} ( x - 1 ) ^{4}}

\frac{\partial}{\partial y} (3 e^{x} sin (y))

\frac{d}{d x} (\frac{1 - x}{x})

\int \frac{1}{x ^{2} - 2 x + 101} d x

\int sec^{5} (4 x) d x

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

What is (\partial)/(\partial x)(ln(1+(x-1)^4*y^2)) ?
The answer to (\partial)/(\partial x)(ln(1+(x-1)^4*y^2)) is (4y^2(x-1)^3)/(1+y^2(x-1)^4)