Question

1. let $y^{4}+xy + 4sqrt{x}=13$. what is $\frac{dy}{dx}$ at $(4,1)$? solution:

Explanation:

Step1: Differentiate both sides

Differentiate $y^{4}+xy + 4\sqrt{x}=13$ with respect to $x$.
Using the chain - rule, product - rule and power - rule:
The derivative of $y^{4}$ with respect to $x$ is $4y^{3}\frac{dy}{dx}$ (by chain - rule: if $u = y$, then $\frac{d}{dx}(u^{4})=4u^{3}\frac{du}{dx}=4y^{3}\frac{dy}{dx}$).
The derivative of $xy$ with respect to $x$ is $y + x\frac{dy}{dx}$ (by product - rule: $(uv)^\prime=u^\prime v+uv^\prime$, where $u = x$, $v = y$, $u^\prime=1$, $v^\prime=\frac{dy}{dx}$).
The derivative of $4\sqrt{x}=4x^{\frac{1}{2}}$ with respect to $x$ is $4\times\frac{1}{2}x^{-\frac{1}{2}}=\frac{2}{\sqrt{x}}$.
The derivative of the constant 13 with respect to $x$ is 0.
So, $4y^{3}\frac{dy}{dx}+y + x\frac{dy}{dx}+\frac{2}{\sqrt{x}}=0$.

Step2: Solve for $\frac{dy}{dx}$

Group the terms with $\frac{dy}{dx}$ on one side:
$4y^{3}\frac{dy}{dx}+x\frac{dy}{dx}=-y - \frac{2}{\sqrt{x}}$.
Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(4y^{3}+x)=-y - \frac{2}{\sqrt{x}}$.
Then $\frac{dy}{dx}=\frac{-y-\frac{2}{\sqrt{x}}}{4y^{3}+x}$.

Step3: Substitute $x = 4$ and $y = 1$

Substitute $x = 4$ and $y = 1$ into the expression for $\frac{dy}{dx}$:
$\frac{dy}{dx}\big|_{(4,1)}=\frac{-1-\frac{2}{\sqrt{4}}}{4\times1^{3}+4}$.
First, simplify the numerator: $-1-\frac{2}{\sqrt{4}}=-1 - 1=-2$.
Then, simplify the denominator: $4\times1^{3}+4=4 + 4 = 8$.
So, $\frac{dy}{dx}\big|_{(4,1)}=-\frac{2}{8}=-\frac{1}{4}$.

Answer:

$-\frac{1}{4}$