# IB μαθηματικά HL- Differentiation

##### Differentiation, Derivatives, gradient, tangent- IB Mathematics HL
1. 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

$$f(x)=x^4+5x^2+3x+4\Rightarrow$$ $$f’(x)=4x^3+10x+3\Rightarrow$$

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

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