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Q. 21

Expert-verified
Found in: Page 976

Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861

In Exercises 21–26, find the discriminant of the given function.$f\left(x,y\right)={e}^{2x}\mathrm{cos}y$.

The answer is $-4{e}^{4x}$.

See the step by step solution

Step 1. Given Information.

The function is $f\left(x,y\right)={e}^{2x}\mathrm{cos}y$.

Step 2. Explanation.

The discriminant is calculated by formula,

$det\left({H}_{f}\right)=\frac{{\partial }^{2}f}{\partial {x}^{2}}·\frac{{\partial }^{2}f}{\partial {y}^{2}}-{\left(\frac{{\partial }^{2}f}{\partial x\partial y}\right)}^{2}$

Find $\frac{\partial f}{\partial x}$, $\frac{{\partial }^{2}f}{\partial {x}^{2}}$, $\frac{\partial f}{\partial y}$, $\frac{{\partial }^{2}f}{\partial {y}^{2}}$ and localid="1649783903210" $\frac{{\partial }^{2}f}{\partial x\partial y}$

$\frac{\partial f}{\partial x}=2{e}^{2x}\mathrm{cos}y$, $\frac{{\partial }^{2}f}{\partial {x}^{2}}=4{e}^{2x}\mathrm{cos}y$

$\frac{\partial f}{\partial y}=-{e}^{2x}\mathrm{sin}y$, localid="1649783850374" $\frac{{\partial }^{2}f}{\partial {y}^{2}}=-{e}^{2x}\mathrm{cos}y$

localid="1649818601411" $\frac{{\partial }^{2}f}{\partial x\partial y}=-2{e}^{2x}\mathrm{sin}y$

Step 3. Calculate Discriminant.

Calculate $det\left({H}_{f}\right)=\frac{{\partial }^{2}f}{\partial {x}^{2}}·\frac{{\partial }^{2}f}{\partial {y}^{2}}-{\left(\frac{{\partial }^{2}f}{\partial x\partial y}\right)}^{2}$

localid="1649818688351" $det\left({H}_{f}\right)=\left(4{e}^{2x}\mathrm{cos}y\right)\left(-{e}^{2x}\mathrm{cos}y\right)-{\left(-2{e}^{2x}\mathrm{sin}y\right)}^{2}\phantom{\rule{0ex}{0ex}}=-4{e}^{4x}{\mathrm{cos}}^{2}y-4{e}^{4x}{\mathrm{sin}}^{2}x\phantom{\rule{0ex}{0ex}}=-4{e}^{4x}$