Question

Calculate Fxxyz if F(x,y,z)=sin(3x+yz) .

Answer 1

Answer on Question #43852 - Math – Differential Calculus | Equations

Calculate $F_{xxyz}$ if $F(x,y,z) = \sin (3x + yz)$.

Solution

Let $F(x,y,z) = \sin (3x + yz)$.

$$F_{xxyz} = \frac{\partial}{\partial z} \frac{\partial}{\partial y} \frac{\partial}{\partial x} \frac{\partial}{\partial z} F(x,y,z) = \frac{\partial}{\partial z} \frac{\partial}{\partial y} \frac{\partial}{\partial x} \frac{\partial}{\partial x} \sin (3x + yz).$$

Calculate step by step four partial derivatives.

First, find the partial derivative of a given function with respect to $x$. To calculate $\frac{\partial}{\partial x} F(x,y,z)$, we simply view $y$ and $z$ as being a fixed number and calculate the ordinary derivative with respect to $x$. In calculating partial derivatives, we can use all the rules for ordinary derivatives.

Recall the derivative of the sine function: $\frac{d}{dx} \sin(x) = \cos(x)$

We can think of continuous function $F(x,y,z) = \sin (3x + yz)$ as being the result of combining two continuous functions. If $g(x) = 3x + yz$ is the inside function and $h(t) = \sin(t)$ is the outside function, then the result of substituting of $g(x)$ into the function $h$ is

$$h(g(x)) = \sin (3x + yz)$$

By the chain rule, if $f(x) = h(g(x))$, then $(f)_x'(x) = (h)_x'(g(x)) \times (g)_x'(x)$.

Here $(\sin (3x + yz))_x' = (\sin)_x'(3x + yz) \times (3x + yz)_x' = \cos (3x + yz) \times (3x + yz)_x'$.

The sum rule tells us that for two functions $u$ and $v$:

$$(u + v)_x'(x) = (u)_x'(x) + (v)_x'(x)$$

Let $u(x) = 3x$, $v(x) = yz$. We view function $v(x) = yz$ being a constant with respect to $x$. The derivative of constant $c$ is zero: $(c)_x' = \frac{d}{dx} c = 0$. Here $(yz)_x' = \frac{\partial}{\partial x} (yz) = 0$.

The derivative of a constant multiplied by a function is the constant multiplied by the derivative of the original function:

$$(af(x))_x' = a(f(x))_x'.$$

Here $(3x)_x' = 3(x)_x' = 3 \times 1 = 3$.

Therefore, $(3x + yz)_x' = (3x)_x' + (yz)_x' = 3 + 0 = 3$.

So,

$$F_x = \frac{\partial}{\partial x} F(x,y,z) = \frac{\partial}{\partial x} \sin (3x + yz) = \left. \frac{d}{dt} \sin (t) \right|_{t=3x+yz} \times (3x + yz)_x' = \cos (3x + yz) \times 3 = 3 \cos (3x + yz).$$

Next, recall the derivative of the cosine function: $\frac{d}{dx}\cos (x) = -\sin (x)$ .

So,

$$\begin{array}{l} F _ {x x} = \frac {\partial}{\partial x} \frac {\partial}{\partial x} F (x, y, z) = \frac {\partial}{\partial x} \frac {\partial}{\partial x} \sin (3 x + y z) = \frac {\partial}{\partial x} \left(\frac {\partial}{\partial x} (\sin (3 x + y z))\right) = \frac {\partial}{\partial x} (3 \cos (3 x + y z)) = \\ = 3 \frac {\partial}{\partial x} (\cos (3 x + y z)) = 3 \frac {d}{d t} \cos (t) \bigg | _ {t = 3 x + y z} \times (3 x + y z) _ {x} ^ {\prime} = - 3 \sin (3 x + y z) \times 3 = \\ \end{array}$$ $$= - 9 \sin (3 x + y z).$$

Next,

$$\begin{array}{l} F _ {x x y} = \frac {\partial}{\partial y} \frac {\partial}{\partial x} \frac {\partial}{\partial x} F (x, y, z) = \frac {\partial}{\partial y} \frac {\partial}{\partial x} \frac {\partial}{\partial x} \sin (3 x + y z) = \frac {\partial}{\partial y} \left(\frac {\partial}{\partial x} \frac {\partial}{\partial x} \sin (3 x + y z)\right) = \frac {\partial}{\partial y} (- 9 \sin (3 x + y z)) = \\ = - 9 \frac {\partial}{\partial y} (\sin (3 x + y z)) = - 9 \frac {d}{d t} \sin (t) \bigg | _ {t = 3 x + y z} \times (3 x + y z) _ {y} ^ {\prime} = - 9 \cos (3 x + y z) \times (0 + z) = \\ = - 9 z \cos (3 x + y z). \\ \end{array}$$

Next,

$$\begin{array}{l} F _ {x x y z} = \frac {\partial}{\partial z} \frac {\partial}{\partial y} \frac {\partial}{\partial x} \frac {\partial}{\partial x} F (x, y, z) = \frac {\partial}{\partial z} \left(\frac {\partial}{\partial y} \frac {\partial}{\partial x} \frac {\partial}{\partial x} F (x, y, z)\right) = \frac {\partial}{\partial z} \left(\frac {\partial}{\partial y} \frac {\partial}{\partial x} \frac {\partial}{\partial x} \sin (3 x + y z)\right) = \\ = \frac {\partial}{\partial z} (- 9 z \cos (3 x + y z)) = - 9 \frac {\partial}{\partial z} (z \cos (3 x + y z)) = - 9 \cos (3 x + y z) \times \frac {\partial z}{\partial z} - 9 z \frac {\partial}{\partial z} (\cos (3 x + y z)) = \\ = - 9 \cos (3 x + y z) \times 1 - 9 z \frac {d}{d t} \cos (t) \bigg | _ {t = 3 x + y z} \times (3 x + y z) _ {z} ^ {\prime} = \\ = - 9 \cos (3 x + y z) - 9 z \times (- \sin (t)) | _ {t = 3 x + y z} \times (0 + y) = \\ = - 9 \cos (3 x + y z) + 9 z \sin (3 x + y z) \times y = \\ \end{array}$$ $$= - 9 \cos (3 x + y z) + 9 y z \sin (3 x + y z).$$

Note, If the mixed partial derivatives exist and are continuous at a point $x_0$ , then they are equal at $x_0$ regardless of the order in which they are taken.

Answer: $F_{xxyz} = -9\cos (3x + yz) + 9yz\sin (3x + yz)$

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