Question

if $3x^{2}+4x + xy=4$ and $y(4)=-15$, find $y(4)$ by implicit differentiation.

Explanation:

Step1: Differentiate both sides

Differentiate $3x^{2}+4x + xy=4$ with respect to $x$.
Using the sum - rule and product - rule:
The derivative of $3x^{2}$ is $6x$ (since $\frac{d}{dx}(ax^{n})=nax^{n - 1}$), the derivative of $4x$ is $4$, and for $xy$ using the product - rule $\frac{d}{dx}(uv)=u'v + uv'$ where $u = x$ and $v = y$, we get $y+xy'$. The derivative of the constant $4$ is $0$.
So, $6x + 4+y+xy'=0$.

Step2: Solve for $y'$

Rearrange the equation $6x + 4+y+xy'=0$ to isolate $y'$.
First, move the non - $y'$ terms to the other side: $xy'=-6x - 4 - y$.
Then, divide by $x$ (assuming $x
eq0$) to get $y'=\frac{-6x - 4 - y}{x}$.

Step3: Substitute $x = 4$ and $y(4)=-15$

Substitute $x = 4$ and $y=-15$ into the formula for $y'$.
$y'(4)=\frac{-6\times4 - 4-(-15)}{4}=\frac{-24 - 4 + 15}{4}=\frac{-13}{4}$.

Answer:

$-\frac{13}{4}$