Question

1. find $y=\frac{dy}{dx}$ for the following curves.

a. $\tan(y)=\frac{x^{3}}{10}$
b. $x^{3}+xy + y^{3}=10$
c. $x^{2}y - e^{y}=x + 1$
d. $cos(xy)=x^{2}+y^{2}$

Explanation:

Step1: Differentiate both sides of A

Differentiate $\tan(y)=\frac{x^{3}}{10}$ with respect to $x$. Using the chain - rule on the left - hand side ($\frac{d}{dx}\tan(y)=\sec^{2}(y)\cdot y'$) and the power rule on the right - hand side ($\frac{d}{dx}\frac{x^{3}}{10}=\frac{3x^{2}}{10}$). So, $\sec^{2}(y)y'=\frac{3x^{2}}{10}$, then $y'=\frac{3x^{2}}{10\sec^{2}(y)}$.

Step2: Differentiate both sides of B

Differentiate $x^{3}+xy + y^{3}=10$ with respect to $x$. Using the power rule, product rule ($(uv)' = u'v+uv'$ where $u = x$ and $v = y$), and chain - rule. $\frac{d}{dx}(x^{3})=3x^{2}$, $\frac{d}{dx}(xy)=y + xy'$, $\frac{d}{dx}(y^{3})=3y^{2}y'$. So, $3x^{2}+y+xy'+3y^{2}y' = 0$. Rearranging for $y'$ gives $y'(x + 3y^{2})=-3x^{2}-y$, and $y'=\frac{-3x^{2}-y}{x + 3y^{2}}$.

Step3: Differentiate both sides of C

Differentiate $x^{2}y-e^{y}=x + 1$ with respect to $x$. Using the product rule on $x^{2}y$ ($\frac{d}{dx}(x^{2}y)=2xy+x^{2}y'$) and the chain - rule on $e^{y}$ ($\frac{d}{dx}(e^{y})=e^{y}y'$). So, $2xy+x^{2}y'-e^{y}y'=1$. Rearranging for $y'$ gives $y'(x^{2}-e^{y})=1 - 2xy$, and $y'=\frac{1 - 2xy}{x^{2}-e^{y}}$.

Step4: Differentiate both sides of D

Differentiate $\cos(xy)=x^{2}+y^{2}$ with respect to $x$. Using the chain - rule on $\cos(xy)$ ($\frac{d}{dx}\cos(xy)=-\sin(xy)\cdot(y+xy')$) and the power rule on the right - hand side. So, $-\sin(xy)(y + xy')=2x+2yy'$. Expanding gives $-y\sin(xy)-x\sin(xy)y'=2x + 2yy'$. Rearranging for $y'$: $-x\sin(xy)y'-2yy'=2x + y\sin(xy)$, $y'(-x\sin(xy)-2y)=2x + y\sin(xy)$, and $y'=\frac{2x + y\sin(xy)}{-x\sin(xy)-2y}$.

Answer:

A. $y'=\frac{3x^{2}}{10\sec^{2}(y)}$
B. $y'=\frac{-3x^{2}-y}{x + 3y^{2}}$
C. $y'=\frac{1 - 2xy}{x^{2}-e^{y}}$
D. $y'=\frac{2x + y\sin(xy)}{-x\sin(xy)-2y}$