Find the equation of the tangent to the curve y = 2 ln(2e - x) at the point on the curve where x = e.
The first step is to find the y coordinate at the point on the curve where x = e. To do this we subsitute x = e into the equation of the curve, so y = 2ln(2e - x) becomes y = 2ln(2e - e). Simplifying this, we get y = 2ln(e). Using our knowledge of the natural logarithm, ln(e) = 1, so therefore y = 2.
Next, we find the gradient of the curve at the point x = e. We do this by differentiating the equation, and then subsituting the value for x in. Note that we have to use the chain rule.
dy/dx = -2/(2e - x) = -2/(2e - e) = -2/e.
Using all the information we've found, we can now produce the equation of the tangent line using the equation of a line formula, y - y0 = m(x - x0), where m is the gradient.
y - 2 = -2/e (x - e). Rearranging this, we get y = 4 - 2x/e as the equation of the tangent line.
Related Maths A Level answers
All answers ▸Find the solutions to x^3+4x^2+x-5=1
Find the gradient of the function f(x,y)=x^3 + y^3 -3xy at the point (2,1), given that f(2,1) = 6.
Consider a differential equation where dx/dt = -axt. Find an equation for x(t).
Find the x-values of the turning points on the graph, y=(3-x)(x^2-2)