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

Solution

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

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

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

Applying chain rule

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

At point (1,2)

$\frac{dy}{dx}=\frac{(2)^2+4(1)(2)^2}{4(1)^2(2)-2(1)(2)}=5$

Using slope intercept equation

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