Answer to Question #277966 in Calculus for Nick

Question #277966

2. a) Find the derivatives of the following functions with respect to x.








x ^ 3 + y ^ 3 = 3






y = (sin x) ^ tan x








b) Evaluate the 2 ^ (nd) order partial derivatives partial^ 2 u partial x^ 2 and partial^ 2 u partial y^ 2 if u=2x^ 3 +3x^ 2 y+xy^ +y^ .

1
Expert's answer
2021-12-14T10:05:04-0500

x3+y3=3y3=3x3Differentiating implicitly, we get3y2dydx=3x2dydx=x2y2y=sin(x)tan(x)We take the natural logarithm to obtainlny=tanxlnsinxNext, we take the exponential of both sides to obtain1ydydx=sec2xlnsinx+tanx1sinxcosx=1+lnsinxsec2x    1ydydx==1+lnsinxsec2xMultiplying both sides by y=sinxtanx, we haved(sinxtanx)dx=(1+lnsinxsec2x)sinxtanxu=2x3+3x2y+xy2+y2The 2nd order partial derivatives is given by uxx,uxy,uyyuxx=12x+6yuxy=6x+2yuyx=6x+2yuyy=2x+2\displaystyle x^3 +y^3 =3\\ y^3 = 3-x^3\\ \text{Differentiating implicitly, we get}\\ 3y^2\frac{dy}{dx}=-3x^2\\ \frac{dy}{dx}= -\frac{x^2}{y^2}\\ y =\sin(x)^{\tan(x)}\\ \text{We take the natural logarithm to obtain}\\ \ln y = \tan x \ln \sin x\\ \text{Next, we take the exponential of both sides to obtain}\\ \frac{1}{y}\cdot\frac{dy}{dx} = \sec^2x\ln \sin x + \tan x \frac{1}{\sin x}\cdot \cos x\\ = 1 + \ln \sin x \sec^2x\\ \implies \frac{1}{y}\cdot\frac{dy}{dx} = = 1 + \ln \sin x \sec^2x\\ \text{Multiplying both sides by $y = \sin x^{\tan x}$, we have}\\ \frac{d(\sin x^{\tan x})}{dx}= (1+ \ln \sin x \sec^2x)\sin x^{\tan x}\\ u = 2x^3 + 3x^2y+xy^2+y^2\\ \text{The 2nd order partial derivatives is given by $u_{xx}, u_{xy}, u_{yy}$}\\ u_{xx}= 12x + 6y\\ u_{xy}= 6x+2y\\ u_{yx} = 6x+2y\\ u_{yy} = 2x + 2


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