Differentiation equations of lines

Question

Find an equation of the tangent line to the curve cos(𝑥)+11𝑦^2=𝑥𝑦^7+34 at the point (0,(3)^1/2). Assume that 𝑦 is a function of 𝑥.

Express all numbers in exact form and write the equation of the tangent line in terms of 𝑥 and 𝑦.

Answer 1

cos(𝑥) + 11𝑦² = 𝑥𝑦⁷ + 34

To find the slope of the tangent line, take the derivative. Since this is complex with x's and y's in multiple locations, take the derivative implicitly:

The derivative of cos(x) with respect to "x" is -sin(x)

The derivative of 11y² with respect to "x" is 22y•y' (using the power rule and the chain rule)

The derivative of xy⁷ requires use of the product rule (u•v)' = u'v + uv' where u = x, u' = 1, v = y⁷, v' = 7y⁶•y' so (xy⁷)' = (1)(y⁷) + (x)(7y⁶•y') = y⁷ + 7xy⁶•y'

The derivative of 34 = 0

Putting these together we get:

-sin(x) + 22y•y' = y⁷ + 7xy⁶•y'

22y•y' - 7xy⁶•y' = y⁷ + sin(x)

y'(22y - 7xy⁶) = y⁷ + sin(x)

y' = (y⁷ + sin(x))/(22y - 7xy⁶)

At the point (0, √3) we plug in x = 0 and y = √3 to get:

y' = ((√3)⁷ + sin(0)/(22√3 - 7(0)(√3)⁶)

y' = (√3)⁶•√3 + 0)/(22√3 - 0)

y' = (27√3)/(22√3)

y' = 27/22

This is the slope of the tangent line. So, using the point-slope form, the equation of the line is:

y - y₁ = m(x - x₁)

y - √3 = (27/22)(x - 0)

y = (27/22)x + √3