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 for 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, $\sec^{2}(y)\cdot y'=\frac{3x^{2}}{10}$, then $y'=\frac{3x^{2}}{10\sec^{2}(y)}$.

Step2: Differentiate both sides for B

Differentiate $x^{3}+xy + y^{3}=10$ with respect to $x$. Using the power - rule, product - rule 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, $3x^{2}+y + xy'+3y^{2}y'=0$. Rearrange to get $y'(x + 3y^{2})=-3x^{2}-y$, then $y'=\frac{-3x^{2}-y}{x + 3y^{2}}$.

Step3: Differentiate both sides for C

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

Step4: Differentiate both sides for D

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

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. $y'=\frac{2x + y\sin(xy)}{-x\sin(xy)-2y}$