tentukan laju perubahan maksimum $f(x, y, z) = \ln \left(\frac{2x+3y}{z}\right)$ di titik $(2, 7, 4)$ dan arah yang menyebabkan laju perubahan maksimum

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Find derivatives using the quotient rule II

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Q. tentukan laju perubahan maksimum

f(x, y, z) = \ln \left(\frac{2x+3y}{z}\right)

di titik

(2, 7, 4)

dan arah yang menyebabkan laju perubahan maksimum

Calculate gradient vector: To find the maximum rate of change, we need to calculate the gradient of $f(x, y, z)$ . The gradient is a vector of partial derivatives.
Find partial derivatives: First, let's find the partial derivative of $f$ with respect to $x$ : $\frac{\partial f}{\partial x} = \frac{1}{z} \cdot \frac{2}{2x+3y}$ .
Calculate at point $(2, 7, 4)$ : At the point $(2, 7, 4)$ , $\frac{\partial f}{\partial x} = \frac{1}{4} \cdot \left(\frac{2}{4+21}\right) = \frac{1}{50}$ .
Find gradient vector: Now, let's find the partial derivative of $f$ with respect to $y$ : $\frac{\partial f}{\partial y} = \left(\frac{1}{z}\right) \cdot \left(\frac{3}{2x+3y}\right)$ .
Determine maximum rate of change: At the point $(2, 7, 4)$ , $\frac{\partial f}{\partial y} = \left(\frac{1}{4}\right) * \left(\frac{3}{4+21}\right) = \frac{3}{100}$ .
Calculate magnitude: Next, we find the partial derivative of $f$ with respect to $z$ : $\frac{\partial f}{\partial z} = -\left(\ln(2x+3y)\right)/z^2$ .
Calculate magnitude: Next, we find the partial derivative of $f$ with respect to $z$ : $\frac{\partial f}{\partial z} = -\frac{\ln(2x+3y)}{z^2}$ .At the point $(2, 7, 4)$ , $\frac{\partial f}{\partial z} = -\frac{\ln(4+21)}{16} = -\frac{\ln(25)}{16}$ .
Calculate magnitude: Next, we find the partial derivative of $f$ with respect to $z$ : $\frac{\partial f}{\partial z} = -\frac{\ln(2x+3y)}{z^2}$ .At the point $(2, 7, 4)$ , $\frac{\partial f}{\partial z} = -\frac{\ln(4+21)}{16} = -\frac{\ln(25)}{16}$ .The gradient vector at the point $(2, 7, 4)$ is $\newlineabla f = \left(\frac{1}{50}, \frac{3}{100}, -\frac{\ln(25)}{16}\right)$.
Calculate magnitude: Next, we find the partial derivative of $f$ with respect to $z$ : $\frac{\partial f}{\partial z} = -\frac{\ln(2x+3y)}{z^2}$ .At the point $(2, 7, 4)$ , $\frac{\partial f}{\partial z} = -\frac{\ln(4+21)}{16} = -\frac{\ln(25)}{16}$ .The gradient vector at the point $(2, 7, 4)$ is $\newlineabla f = \left(\frac{1}{50}, \frac{3}{100}, -\frac{\ln(25)}{16}\right)$.The maximum rate of change occurs in the direction of the gradient vector. So, we need to find the magnitude of the gradient vector to get the maximum rate of change.
Calculate magnitude: Next, we find the partial derivative of $f$ with respect to $z$ : $\frac{\partial f}{\partial z} = -\frac{\ln(2x+3y)}{z^2}$ .At the point $(2, 7, 4)$ , $\frac{\partial f}{\partial z} = -\frac{\ln(4+21)}{16} = -\frac{\ln(25)}{16}$ .The gradient vector at the point $(2, 7, 4)$ is $\newlineabla f = \left(\frac{1}{50}, \frac{3}{100}, -\frac{\ln(25)}{16}\right)$.The maximum rate of change occurs in the direction of the gradient vector. So, we need to find the magnitude of the gradient vector to get the maximum rate of change.The magnitude of the gradient vector is |$\newlineabla f| = \sqrt{\left(\frac{1}{50}\right)^2 + \left(\frac{3}{100}\right)^2 + \left(-\frac{\ln(25)}{16}\right)^2}$.
Calculate magnitude: Next, we find the partial derivative of $f$ with respect to $z$ : $\frac{\partial f}{\partial z} = -\frac{\ln(2x+3y)}{z^2}$ .At the point $(2, 7, 4)$ , $\frac{\partial f}{\partial z} = -\frac{\ln(4+21)}{16} = -\frac{\ln(25)}{16}$ .The gradient vector at the point $(2, 7, 4)$ is $\newlineabla f = \left(\frac{1}{50}, \frac{3}{100}, -\frac{\ln(25)}{16}\right)$.The maximum rate of change occurs in the direction of the gradient vector. So, we need to find the magnitude of the gradient vector to get the maximum rate of change.The magnitude of the gradient vector is |$\newlineabla f| = \sqrt{\left(\frac{1}{50}\right)^2 + \left(\frac{3}{100}\right)^2 + \left(-\frac{\ln(25)}{16}\right)^2}$.Calculating the magnitude: |$\newlineabla f| = \sqrt{\left(\frac{1}{2500}\right) + \left(\frac{9}{10000}\right) + \left(\frac{\ln(25)^2}{256}\right)}$.
Calculate magnitude: Next, we find the partial derivative of $f$ with respect to $z$ : $\frac{\partial f}{\partial z} = -\frac{\ln(2x+3y)}{z^2}$ .At the point $(2, 7, 4)$ , $\frac{\partial f}{\partial z} = -\frac{\ln(4+21)}{16} = -\frac{\ln(25)}{16}$ .The gradient vector at the point $(2, 7, 4)$ is $\newlineabla f = \left(\frac{1}{50}, \frac{3}{100}, -\frac{\ln(25)}{16}\right)$.The maximum rate of change occurs in the direction of the gradient vector. So, we need to find the magnitude of the gradient vector to get the maximum rate of change.The magnitude of the gradient vector is |$\newlineabla f| = \sqrt{\left(\frac{1}{50}\right)^2 + \left(\frac{3}{100}\right)^2 + \left(-\frac{\ln(25)}{16}\right)^2}$.Calculating the magnitude: |$\newlineabla f| = \sqrt{\left(\frac{1}{2500}\right) + \left(\frac{9}{10000}\right) + \left(\frac{\ln(25)^2}{256}\right)}$.Simplifying the magnitude: |$\newlineabla f| = \sqrt{\left(\frac{1}{2500}\right) + \left(\frac{9}{10000}\right) + \left(\frac{\ln(5)^4}{256}\right)}$.

tentukan laju perubahan maksimum f(x,y,z)=ln⁡(2x+3yz)f(x, y, z) = \ln \left(\frac{2x+3y}{z}\right)f(x,y,z)=ln(z2x+3y​) di titik (2,7,4)(2, 7, 4)(2,7,4) dan arah yang menyebabkan laju perubahan maksimum

tentukan laju perubahan maksimum $f(x, y, z) = \ln \left(\frac{2x+3y}{z}\right)$ di titik $(2, 7, 4)$ dan arah yang menyebabkan laju perubahan maksimum