given the equation below, find \\( \\frac { d y } { d x } \\).\n\\( 18 x ^ { 10 } + 2 x ^ { 45 } y + y ^ { 7…

Explanation:

Step1: Differentiate both sides with respect to (x)

Differentiate (18x^{10}+2x^{45}y + y^{7}=21) term - by - term. Using the power rule (\frac{d}{dx}(x^n)=nx^{n - 1}), product rule (\frac{d}{dx}(uv)=u'v + uv') (where (u = 2x^{45}), (v=y)), and chain rule (\frac{d}{dx}(y^n)=ny^{n - 1}\frac{dy}{dx}). (\frac{d}{dx}(18x^{10})=18\times10x^{9}=180x^{9}) (\frac{d}{dx}(2x^{45}y)=2\times45x^{44}y+2x^{45}\frac{dy}{dx}=90x^{44}y + 2x^{45}\frac{dy}{dx}) (\frac{d}{dx}(y^{7})=7y^{6}\frac{dy}{dx}) (\frac{d}{dx}(21) = 0) So, (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}): (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}})

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

Substitute (x = 1) and (y = 1) into (\frac{dy}{dx}): (m=\frac{dy}{dx}\big|_{x = 1,y = 1}=\frac{-180\times1^{9}-90\times1^{44}\times1}{2\times1^{45}+7\times1^{6}}=\frac{-180 - 90}{2 + 7}=\frac{-270}{9}=-30)

Step4: Use the point - slope form (y - y_1=m(x - x_1)) to find the tangent line

Given (m=-30), (x_1 = 1), (y_1 = 1) (y-1=-30(x - 1)) Expand: (y-1=-30x + 30) (y=-30x+31)

Answer:

(\frac{dy}{dx}=\frac{-180x^{9}-90x^{44}y}{2x^{45}+7y^{6}}) (y=-30x + 31)