Question

The equation of a curve is xy squared minus 2x square y squared equal 0. Find the gradient of the tangent to the curve at (1,2)?

Answer 1

`xy^2-2x^2y^2=0`

we need to find the derivative with respect to x. dy/dx

`\begin{gathered} (d)/(dx)(xy^2-2x^2y^2=0)=(d)/(dx)(xy^2)-2(d)/(dx)(x^2y^2)=0 \\ \\ \text{ using product rule} \\ \\ \Rightarrow x(d)/(dx)(y^2)+y^2(d)/(dx)(x)-2x^2(d)/(dx)(y^2)+2y^2(d)/(dx)(x^2)=0 \\ \end{gathered}`

Applying chain rule

`\begin{gathered} x\cdot(d)/(dy)(y^2)\cdot(dy)/(dx)+y^2-2x^2\cdot(d)/(dy)(y^2)\cdot(dy)/(dx)+2y^2(2x)=0 \\ \\ \Rightarrow x(2y)(dy)/(dx)+y^2-2x^2(2y)(dy)/(dx)+4xy^2=0 \\ \\ \Rightarrow2xy(dy)/(dx)+y^2-4x^2y(dy)/(dx)+4xy^2=0 \\ \\ \Rightarrow(2xy-4x^2y)(dy)/(dx)=-y^2-4xy^2 \\ \\ \Rightarrow(dy)/(dx)=(-(y^2+4xy^2))/(2xy-4x^2y)=(y^2+4xy^2)/(4x^2y-2xy) \\ \\ \Rightarrow(dy)/(dx)=(y^(2)+4xy^(2))/(4x^(2)y-2xy) \end{gathered}`

At point (1,2)

`(dy)/(dx)=((2)^2+4(1)(2)^2)/(4(1)^2(2)-2(1)(2))=5`

Using slope intercept equation

`\begin{gathered} y-y_1=m(x-x_1) \\ \\ y-2=5(x-1) \\ \\ y-2=5x-5 \\ \\ \Rightarrow y=5x-5+2 \\ \\ \Rightarrow y=5x-3 \end{gathered}`