given the equation below, find $\\frac{dy}{dx}$. $18x^{10}+2x^{45}y + y^{7}=21$ $\\frac{dy}{dx}=\\frac{-180x^…

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}})