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

解答

\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}}

求解步骤

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

将

y

视为常数

使用连锁律:

\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}

化简

\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})

d er i v a t i v e y = sin (tan (4) x^{5})

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

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