# IB μαθηματικά SL – Differentiation

##### Differentiation, Derivatives, gradient, tangent – IB Mathematics SL

Find the equation of the tangent to the curve with equation

$$f(x)=2x^3+5x^2-20x+1$$ at the point $$(0,1)$$

Solution

$$f(x)=2x^3+5x^2-20x+1 \Rightarrow$$ $$f’(x)=6x^2+10x-20$$

The gradient of the curve at $$(0,1)$$ is

$$f’(0)=6(0)^2+10(0)-20=-20$$

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

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

where $$m_{t}=-20$$ is the gradient of the tangent

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

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

$$y-1=-20(x-0) \Rightarrow y=-20x+1$$