##### Differentiation, Derivatives, gradient, tangent- IB Mathematics HL

- Find the equation of the tangent to the curve with equation

\( 2x^2+5xy-4y^2-22x=-60\) at the point \((5,0)\)

**Solution**

\( 2x^2+5xy-4y^2-22x=-60 \Rightarrow \) \(\frac{d}{dx}{ 2x^2+5xy-4y^2-22x}=\frac{d}{dx}-60 \Rightarrow \) \(4x+5(y+x \frac{dy}{dx})-8y \frac{dy}{dx}-22=0 \Rightarrow \) \(4x+5y+5x \frac{dy}{dx}-8y \frac{dy}{dx}-22=0 \Rightarrow \) \( \frac{dy}{dx}(5x-8y)=22-4x-5y \Rightarrow \) \( \frac{dy}{dx}=\frac{22-4x-5y}{5x-8y}\Rightarrow \)

The gradient of the curve at \((5,0)\) is

\( \frac{dy}{dx}=\frac{22-4(5)-5(0)}{5(5)-8(0)}\Rightarrow \) \( \frac{dy}{dx}=\frac{2}{25}\)

The equation of the tangent on this point is given by the equation

\(y-y_{0}=m_{t}(x-x_{0})\)

where \( m_{t}=\frac{2}{25} \) is the gradient of the tangent

and the touch point is \((x_{0},y_{0})=(5,0)\).

So, the equation of the tangent at the given point is

\(y=\frac{2}{25} (x-5)=\frac{2}{25} x-\frac{2}{5} \)

2. Find the equation of the tangent to the curve with equation

\( f(x)=x^4+5x^2+3x+4\) at the point \((1,13)\)?

**Solution**

The gradient of the curve at \((1,13)\) is

\(f’(1)=4(1)^3+10(1)+3=17 \)The equation of the tangent on this point is given by the equation

\(y-y_{0}=m_{t}(x-x_{0})\)where \( m_{t}=17 \) is the gradient of the tangent

and the touch point is \((x_{0},y_{0})=(1,13)\).

So, the equation of the tangent at the given point is

\(y-13=17(x-1) \Rightarrow y=17x-4\)