Answer to Question #275027 in Calculus for Asif

Question

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^ .

Answer 1

 $\begin{aligned} &x^{3}+y^{3}=3 \\ &y^{3}=3-x^{3} \end{aligned}$

Differentiating implicitly, we get

$\begin{aligned} &3 y^{2} \frac{d y}{d x}=-3 x^{2} \\ &\frac{d y}{d x}=-\frac{x^{2}}{y^{2}} \\ &y=\sin (x)^{\tan (x)} \end{aligned}$

We take the natural logarithm to obtain $\ln y=\tan x \ln \sin x$

Next, we take the exponential of both sides to obtain

$\begin{aligned} &\frac{1}{y} \cdot \frac{d y}{d x}=\sec ^{2} x \ln \sin x+\tan x \frac{1}{\sin x} \cdot \cos x \\ &=1+\ln \sin x \sec ^{2} x \end{aligned}$

  $\Longrightarrow \frac{1}{y} \cdot \frac{d y}{d x}=1+\ln \sin x \sec ^{2} x$

Multiplying both sides by $y=\sin x^{\tan x}$ , we have

$\begin{aligned} &\frac{d\left(\sin x^{\tan x}\right)}{d x}=\left(1+\ln \sin x \sec ^{2} x\right) \sin x^{\tan x} \\ &u=2 x^{3}+3 x^{2} y+x y^{2}+y^{2}\end{aligned}$

The 2nd order partial derivatives is given by $u_{x x}, u_{x y}, u_{y y}$

$\begin{aligned} &u_{x x}=12 x+6 y \\ &u_{x y}=6 x+2 y \\ &u_{y x}=6 x+2 y \\ &u_{y y}=2 x+2 \end{aligned}$