Question

given the equation below, find $\frac{dy}{dx}$. $17x^{3}+10x^{51}y + y^{7}=28$ $\frac{dy}{dx}=$ now, find the equation of the tangent line to the curve at $(1, 1)$. write your answer in $mx + b$ format $y=$

Explanation:

Step1: Differentiate each term

Differentiate $17x^{3}+10x^{51}y + y^{7}=28$ term - by - term with respect to $x$.
The derivative of $17x^{3}$ using the power rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$ is $51x^{2}$.
For the term $10x^{51}y$, use the product rule $\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}$, where $u = 10x^{51}$ and $v = y$. So its derivative is $10\times51x^{50}y+10x^{51}\frac{dy}{dx}=510x^{50}y + 10x^{51}\frac{dy}{dx}$.
The derivative of $y^{7}$ with respect to $x$ using the chain - rule $\frac{d}{dx}(y^{n})=ny^{n - 1}\frac{dy}{dx}$ is $7y^{6}\frac{dy}{dx}$.
The derivative of the constant 28 is 0. So we have:
$51x^{2}+510x^{50}y + 10x^{51}\frac{dy}{dx}+7y^{6}\frac{dy}{dx}=0$.

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

Group the terms with $\frac{dy}{dx}$ on one side:
$10x^{51}\frac{dy}{dx}+7y^{6}\frac{dy}{dx}=-51x^{2}-510x^{50}y$.
Factor out $\frac{dy}{dx}$:
$\frac{dy}{dx}(10x^{51}+7y^{6})=-51x^{2}-510x^{50}y$.
Then $\frac{dy}{dx}=\frac{-51x^{2}-510x^{50}y}{10x^{51}+7y^{6}}$.

Step3: Find the slope at the point $(1,1)$

Substitute $x = 1$ and $y = 1$ into $\frac{dy}{dx}$:
$\frac{dy}{dx}\big|_{(1,1)}=\frac{-51\times1^{2}-510\times1^{50}\times1}{10\times1^{51}+7\times1^{6}}=\frac{-51 - 510}{10 + 7}=\frac{-561}{17}=-33$.

Step4: Find the equation of the tangent line

Use the point - slope form of a line $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(1,1)$ and $m=-33$.
$y - 1=-33(x - 1)$.
Expand to get $y-1=-33x + 33$.
Rearrange to the $y=mx + b$ form: $y=-33x+34$.

Answer:

$\frac{dy}{dx}=\frac{-51x^{2}-510x^{50}y}{10x^{51}+7y^{6}}$
$y=-33x + 34$