Find the indicated partial derivative.
$$u = x ^ { a } y ^ { b } z ^ { c } ; \frac { \partial ^ { 6 } u } { \partial x \partial y ^ { 2 } \partial z ^ { 3 } } , a 1 , b 2 , c 3$$
A)
$\frac { \partial ^ { 6 } u } { \partial x \partial y ^ { 2 } \partial z ^ { 3 } } = c b ( b - 1 ) c ( a - 1 ) ( a - 2 ) x ^ { c - 1 } y ^ { b - 2 } z ^ { a - 3 }$
B) $\frac { \partial ^ { 6 } u } { \partial x \partial y ^ { 2 } \partial z ^ { 3 } } = x ^ { a - 1 } y ^ { 3 - 2 } z ^ { c - 3 }$
C)
$\frac { \partial ^ { 6 } u } { \partial x \partial y ^ { 2 } \partial z ^ { 3 } } = a b ( b - 1 ) c ( c - 1 ) ( c - 2 ) x ^ { a - 1 } y ^ { b - 2 } z ^ { c - 3 }$
D)
$\frac { \partial ^ { 6 } u } { \partial x \partial y ^ { 2 } \partial z ^ { 3 } } = a c b ( a - 1 ) ( a - 2 ) x ^ { a - 1 } y ^ { b - 2 } z ^ { c - 3 }$
E)
$\frac { \partial ^ { 6 } u } { \partial x \partial y ^ { 2 } \partial z ^ { 3 } } = x ^ { b - 1 } y ^ { c - 2 } z ^ { a - 3 }$

Question

Find the indicated partial derivative.
$$u = x ^ { a } y ^ { b } z ^ { c } ; \frac { \partial ^ { 6 } u } { \partial x \partial y ^ { 2 } \partial z ^ { 3 } } , a > 1 , b > 2 , c > 3$$
A)
$\frac { \partial ^ { 6 } u } { \partial x \partial y ^ { 2 } \partial z ^ { 3 } } = c b ( b - 1 ) c ( a - 1 ) ( a - 2 ) x ^ { c - 1 } y ^ { b - 2 } z ^ { a - 3 }$
B) $\frac { \partial ^ { 6 } u } { \partial x \partial y ^ { 2 } \partial z ^ { 3 } } = x ^ { a - 1 } y ^ { 3 - 2 } z ^ { c - 3 }$
C)
$\frac { \partial ^ { 6 } u } { \partial x \partial y ^ { 2 } \partial z ^ { 3 } } = a b ( b - 1 ) c ( c - 1 ) ( c - 2 ) x ^ { a - 1 } y ^ { b - 2 } z ^ { c - 3 }$
D)
$\frac { \partial ^ { 6 } u } { \partial x \partial y ^ { 2 } \partial z ^ { 3 } } = a c b ( a - 1 ) ( a - 2 ) x ^ { a - 1 } y ^ { b - 2 } z ^ { c - 3 }$
E)
$\frac { \partial ^ { 6 } u } { \partial x \partial y ^ { 2 } \partial z ^ { 3 } } = x ^ { b - 1 } y ^ { c - 2 } z ^ { a - 3 }$

Quizplus · Accepted Answer

We can find the sixth partial derivative by differentiating the function repeatedly with respect to x, y, and z in that order.

Firstly, differentiating partially with respect to x 6 times will reduce the power of x by 6, giving:

$$ \frac{\partial^6 u}{\partial x^6} = a(a-1)(a-2)(a-3)(a-4)(a-5) x^{a-6}y^b z^c $$

Next, differentiating partially with respect to $y$ twice will reduce the power of $y$ by 2, giving:

$$ \frac{\partial^6 u}{\partial x\partial y^2\partial z^3} = a(a-1)(a-2)(a-3)(a-4)(a-5) x^{a-6}y^{b-2}z^{c-3}\frac{\partial^4 u}{\partial x^4} $$

Now we need to find the fourth partial derivative with respect to $x$. This will reduce the power of $x$ by 4, giving:

$$ \frac{\partial^4 u}{\partial x^4} = a(a-1)(a-2)(a-3) x^{a-4}y^b z^c $$

Finally, combining all the parts, we get:

$$ \frac{\partial^6 u}{\partial x\partial y^2\partial z^3} = a(a-1)(a-2)(a-3)(a-4)(a-5) x^{a-6}y^{b-2}z^{c-3}a(a-1)(a-2)(a-3)x^{a-4}y^b z^c $$

$$ = \mathbf{a b ( b - 1 ) c ( c - 1 ) ( c - 2 ) x ^ { a - 1 } y ^ { b - 2 } z ^ { c - 3 }} $$

Therefore, the best choice is C.

Find the Indicated Partial Derivative u=xaybzc;∂6u∂x∂y2∂z3,a>1,b>2,c>3u = x ^ { a } y ^ { b } z ^ { c } ; \frac { \partial ^ { 6 } u } { \partial x \partial y ^ { 2 } \partial z ^ { 3 } } , a > 1 , b > 2 , c > 3u=xaybzc;∂x∂y2∂z3∂6u​,a>1,b>2,c>3

Find the Indicated Partial Derivative $u = x ^ { a } y ^ { b } z ^ { c } ; \frac { \partial ^ { 6 } u } { \partial x \partial y ^ { 2 } \partial z ^ { 3 } } , a > 1 , b > 2 , c > 3$