If $f(x, y)=x y$ , find the gradient vector $\newlineabla f(6,5) $. $\newline$ $abla f(6,5)=\langle 5,6\rangle$ $\newline$ Use the gradient vector to find the tangent line to the level curve $f(x, y)=30$ at the point $(6,5)$ .

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Find derivatives using logarithmic differentiation

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Q. If

f(x, y)=x y

, find the gradient vector $\newlineabla f(6,5) $.

\newline

abla f(6,5)=\langle 5,6\rangle

\newline

Use the gradient vector to find the tangent line to the level curve

f(x, y)=30

at the point

(6,5)

Calculate Partial Derivatives: To find the gradient vector of the function $f(x, y) = xy$ at the point $(6,5)$ , we need to calculate the partial derivatives of $f$ with respect to $x$ and $y$ .
Evaluate at Point: The partial derivative of $f$ with respect to $x$ is $f_x(x, y) = y$ , because when we differentiate $xy$ with respect to $x$ , $y$ is treated as a constant.
Find Gradient Vector: The partial derivative of $f$ with respect to $y$ is $f_y(x, y) = x$ , because when we differentiate $xy$ with respect to $y$ , $x$ is treated as a constant.
Use Gradient for Tangent Line: Now we evaluate the partial derivatives at the point $(6,5)$ . So $f_x(6, 5) = 5$ and $f_y(6, 5) = 6$ .
Write Tangent Line Equation: The gradient vector at the point $(6,5)$ is therefore $\text{grad} f(6,5) = (5, 6)$ .
Write Tangent Line Equation: The gradient vector at the point $(6,5)$ is therefore $\text{grad} f(6,5) = (5, 6)$ .To find the equation of the tangent line to the level curve $f(x, y) = 30$ at the point $(6,5)$ , we use the gradient vector as the direction vector of the tangent line.
Write Tangent Line Equation: The gradient vector at the point $(6,5)$ is therefore $\text{grad} f(6,5) = (5, 6)$ .To find the equation of the tangent line to the level curve $f(x, y) = 30$ at the point $(6,5)$ , we use the gradient vector as the direction vector of the tangent line.The equation of the tangent line can be written in the form $y - y_1 = m(x - x_1)$ , where $m$ is the slope of the tangent line and $(x_1, y_1)$ is the point of tangency.
Write Tangent Line Equation: The gradient vector at the point $(6,5)$ is therefore $\newlineabla f(6,5) = (5, 6)$.To find the equation of the tangent line to the level curve $f(x, y) = 30$ at the point $(6,5)$ , we use the gradient vector as the direction vector of the tangent line.The equation of the tangent line can be written in the form $y - y_1 = m(x - x_1)$ , where $m$ is the slope of the tangent line and $(x_1, y_1)$ is the point of tangency.Since the gradient vector is perpendicular to the level curve, the slope of the tangent line is the negative reciprocal of the slope represented by the gradient vector. However, in this case, we are dealing with a multivariable function, so we use the gradient vector directly to write the equation of the tangent line.
Write Tangent Line Equation: The gradient vector at the point $(6,5)$ is therefore $\newlineabla f(6,5) = (5, 6)$.To find the equation of the tangent line to the level curve $f(x, y) = 30$ at the point $(6,5)$ , we use the gradient vector as the direction vector of the tangent line.The equation of the tangent line can be written in the form $y - y_1 = m(x - x_1)$ , where $m$ is the slope of the tangent line and $(x_1, y_1)$ is the point of tangency.Since the gradient vector is perpendicular to the level curve, the slope of the tangent line is the negative reciprocal of the slope represented by the gradient vector. However, in this case, we are dealing with a multivariable function, so we use the gradient vector directly to write the equation of the tangent line.The equation of the tangent line in point-slope form is $(y - 5) = (\frac{6}{5})(x - 6)$ , using the components of the gradient vector as the slope.
Write Tangent Line Equation: The gradient vector at the point $(6,5)$ is therefore $\text{grad} f(6,5) = (5, 6)$ .To find the equation of the tangent line to the level curve $f(x, y) = 30$ at the point $(6,5)$ , we use the gradient vector as the direction vector of the tangent line.The equation of the tangent line can be written in the form $y - y_1 = m(x - x_1)$ , where $m$ is the slope of the tangent line and $(x_1, y_1)$ is the point of tangency.Since the gradient vector is perpendicular to the level curve, the slope of the tangent line is the negative reciprocal of the slope represented by the gradient vector. However, in this case, we are dealing with a multivariable function, so we use the gradient vector directly to write the equation of the tangent line.The equation of the tangent line in point-slope form is $(y - 5) = (6/5)(x - 6)$ , using the components of the gradient vector as the slope.Simplifying the equation of the tangent line, we get $y - 5 = (6/5)x - (6/5)\cdot6$ .
Write Tangent Line Equation: The gradient vector at the point $(6,5)$ is therefore $\text{grad} f(6,5) = (5, 6)$ .To find the equation of the tangent line to the level curve $f(x, y) = 30$ at the point $(6,5)$ , we use the gradient vector as the direction vector of the tangent line.The equation of the tangent line can be written in the form $y - y_1 = m(x - x_1)$ , where $m$ is the slope of the tangent line and $(x_1, y_1)$ is the point of tangency.Since the gradient vector is perpendicular to the level curve, the slope of the tangent line is the negative reciprocal of the slope represented by the gradient vector. However, in this case, we are dealing with a multivariable function, so we use the gradient vector directly to write the equation of the tangent line.The equation of the tangent line in point-slope form is $(y - 5) = (6/5)(x - 6)$ , using the components of the gradient vector as the slope.Simplifying the equation of the tangent line, we get $y - 5 = (6/5)x - (6/5)\cdot6$ .Further simplifying, we get $y = (6/5)x - 36/5 + 25/5$ .
Write Tangent Line Equation: The gradient vector at the point $(6,5)$ is therefore $\text{grad} f(6,5) = (5, 6)$ .To find the equation of the tangent line to the level curve $f(x, y) = 30$ at the point $(6,5)$ , we use the gradient vector as the direction vector of the tangent line.The equation of the tangent line can be written in the form $y - y_1 = m(x - x_1)$ , where $m$ is the slope of the tangent line and $(x_1, y_1)$ is the point of tangency.Since the gradient vector is perpendicular to the level curve, the slope of the tangent line is the negative reciprocal of the slope represented by the gradient vector. However, in this case, we are dealing with a multivariable function, so we use the gradient vector directly to write the equation of the tangent line.The equation of the tangent line in point-slope form is $(y - 5) = (6/5)(x - 6)$ , using the components of the gradient vector as the slope.Simplifying the equation of the tangent line, we get $y - 5 = (6/5)x - (6/5)\cdot6$ .Further simplifying, we get $y = (6/5)x - 36/5 + 25/5$ .Finally, we have $\text{grad} f(6,5) = (5, 6)$ $0$ as the equation of the tangent line to the level curve $f(x, y) = 30$ at the point $(6,5)$ .

If $f(x, y)=x y$ , find the gradient vector \(\newlineabla f(6,5) \). $\newline$ $abla f(6,5)=\langle 5,6\rangle$ $\newline$ Use the gradient vector to find the tangent line to the level curve $f(x, y)=30$ at the point $(6,5)$ .

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