Solved: "Find the directional derivative of at a given..."

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2022-01-04

Find the directional derivative of at a given point in the direction indicated by the angle theta. F (x,y)=x^3y^4+x^4y^3, (1,1)

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Skilled2022-02-09Added 403 answers

$F (x, y) = x^{3} y^{4} + x^{4} y^{3}$

By the Sum Rule, the derivative of $x^{3} y^{4} + x^{4} y^{3}$ with respect to $x$ is

$\frac{d}{d x} [x^{3} y^{4}] + \frac{d}{d x} [x^{4} y^{3}]$

Evaluate $\frac{d}{d x} [x^{3} y^{4}]$

Since $y^{4}$ is constant with respect to $x$ , the derivative of $x^{3} y^{4}$ with respect to $x$ is $y^{4} \frac{d}{d x} [x^{3}]$ .

$y^{4} \frac{d}{d x} [x^{3}] + \frac{d}{d x} [x^{4} y^{3}]$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$

$y^{4} (3 x^{2}) + \frac{d}{d x} [x^{4} y^{3}]$

Move $3$ to the left of $y^{4}$

$3 y^{4} x^{2} + \frac{d}{d x} [x^{4} y^{3}]$

Evaluate $\frac{d}{d x} [x^{4} y^{3}]$

Since $y^{3}$ is constant with respect to $x$ , the derivative of $x^{4} y^{3}$ with respect to $x$ is $y^{3} \frac{d}{d x} [x^{4}]$

$3 y^{4} x^{2} + y^{3} \frac{d}{d x} [x^{4}]$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$

$3 y^{4} x^{2} + y^{3} (4 x^{3})$

Move $4$ to the left of $y^{3}$

$3 y^{4} x^{2} + 4 y^{3} x^{3}$

Then put $(1, 1)$ into derivative, where

$x = 1$ and $y = 1$

$3 \times 1^{4} \times 1^{2} + 4 \times 1^{3} \times 1^{3}$

$= 3 + 4$

$= 7$ - Answer

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