[Derivada] Com duas variáveis e derivada mista

por **leticiaeverson** » Dom Abr 22, 2018 00:39

Calcule (df/dx ; df/dy) ,e as derivadas mistas em cada caso:
a) f(x,y)= $3x^{4}$ – 2xy² + y{5}

b) f(x,y)= cos²(3x) + sen²(3y)
c) f(x,y)=Ln(3x –y³)

por **Gebe** » Dom Abr 22, 2018 03:39

leticiaeverson escreveu:Calcule (df/dx ; df/dy) ,e as derivadas mistas em cada caso:
a) f(x,y)= $3x^{4}$ – 2xy² +
b) f(x,y)= cos²(3x) + sen²(3y)
c) f(x,y)=Ln(3x –y³)

a)
$\\ f(x,y)=3x^{4}-2xy^2+y^5\\ \\ \\ \frac{\partial f}{\partial x}=4*3x^{3}-1*2*1y^{2}+0=12x^3-2y^2\\ \\ \\ \frac{\partial f}{\partial y}=0-2*2xy^1+5y^4=-4xy+5y^4\\ \\ \\ \frac{\partial^2 f}{\partial x \partial y}=\frac{\partial \left( \frac{\partial f}{\partial x} \right)}{\partial y}=\frac{\partial \left(12x^3-2y^2 \right)}{\partial y}=0-2*2y^1=-4y$

b)
$\\ f(x,y)=cos^2\left(3x \right)+sen^2\left(3y \right)\\ \\ usando\;a\;regra\;da\;cadeia\\ \frac{\partial f}{\partial x}=2*cos^1(3x)* \left( -sen(3x) \right)*3+2sen^1(3y)*\left(cos(3y) \right)*0=-6cos(3x)sen(3x)\\ \\ \\ \frac{\partial f}{\partial y}=2*cos^1(3x)* \left( -sen(3x) \right)*0+2sen^1(3y)*\left(cos(3y) \right)*3=6cos(3y)sen(3y)\\ \\ \\ \frac{\partial^2 f}{\partial x \partial y}=\frac{\partial \left( \frac{\partial f}{\partial x} \right)}{\partial y}=\frac{\partial \left(-6cos(3x)sen(3x) \right)}{\partial y}=0$

c)
$\\ f(x,y)=ln\left(3x-y^3 \right)\\ \\ usando\;a\;regra\;da\;cadeia\\ \frac{\partial f}{\partial x}=\frac{1}{\left( 3x-y^3 \right)}*\left(3-0 \right)=\frac{3}{3x-y^3}\\ \\ \\ \frac{\partial f}{\partial y}=\frac{1}{\left( 3x-y^3 \right)}*\left(0-3y^2 \right)=-\frac{3y^2}{3x-y^3}\\ \\ \\ \frac{\partial^2 f}{\partial x \partial y}=\frac{\partial \left( \frac{\partial f}{\partial x} \right)}{\partial y}=\frac{\partial \left(\frac{3}{3x-y^3} \right)}{\partial y}=\frac{0*\left(3x-y^3 \right)-\left( -3y^2 \right)*(3)}{\left(3x-y^3 \right)^2}=\frac{9y^2}{\left(3x-y^3 \right)^2}$

Espero ter ajudado. Se ficarem duvidas deixe msg.