Question

given the equation below, find $\frac{dy}{dx}$. $18x^{10}+2x^{45}y + y^{7}=21$ $\frac{dy}{dx}=\frac{-180x^{9}-9044y}{2x^{45}+7y^{6}}$ now, find the equation of the tangent line to the curve at $(1,1)$. write your answer in $mx + b$ format $y=-30x + 31$ question help: video message instructor submit question

Explanation:

Step1: Differentiate each term with respect to \(x\)

Differentiate \(18x^{10}\) using the power rule \((x^n)^\prime = nx^{n - 1}\), we get \(18\times10x^{9}=180x^{9}\).
For \(2x^{45}y\), use the product rule \((uv)^\prime=u^\prime v + uv^\prime\) where \(u = 2x^{45}\), \(u^\prime=2\times45x^{44}=90x^{44}\) and \(v = y\), \(v^\prime=\frac{dy}{dx}\). So the derivative of \(2x^{45}y\) is \(90x^{44}y+2x^{45}\frac{dy}{dx}\).
Differentiate \(y^{7}\) using the chain rule \((y^n)^\prime=ny^{n - 1}\frac{dy}{dx}\), we get \(7y^{6}\frac{dy}{dx}\).
Differentiate the constant \(21\) gives \(0\).
So the derivative of the entire equation \(18x^{10}+2x^{45}y + y^{7}=21\) is:
\(180x^{9}+90x^{44}y+2x^{45}\frac{dy}{dx}+7y^{6}\frac{dy}{dx}=0\)

Step2: Solve for \(\frac{dy}{dx}\)

Group the terms with \(\frac{dy}{dx}\) together:
\(2x^{45}\frac{dy}{dx}+7y^{6}\frac{dy}{dx}=-180x^{9}-90x^{44}y\)
Factor out \(\frac{dy}{dx}\):
\(\frac{dy}{dx}(2x^{45}+7y^{6})=-180x^{9}-90x^{44}y\)
Then \(\frac{dy}{dx}=\frac{-180x^{9}-90x^{44}y}{2x^{45}+7y^{6}}\)

Answer:

\(\frac{dy}{dx}=\frac{-180x^{9}-90x^{44}y}{2x^{45}+7y^{6}}\)