Find the equation of the tangent to the curve y = exp(x) at the point ( a, exp(a) ). Deduce the equation of the tangent to the curve which passes through the point (0,1) .

The tangent to the curve is a straight line and will hence have the form f(x) = m * x + c, where m is the gradient of the tangent line and c is the y-intercept of the tangent line.
The gradient of the tangent to the curve y = exp(x) at point ( a, exp(a) ) may be found by differentiating the curve and evaluating the derivative at point ( a, exp(a) ). Therefore m = dy/dx at x=a = exp(a) .
The equation of the tangent line is now f(x) = exp(a) * x + c, where c is still unknown. We know the tangent line passes through the point ( a, exp(a) ), so substituting these values into the equation of the tangent line gives an expression for c : exp(a) = exp(a) * a + c => c = exp(a) * (1 - a).
The equation of the tangent line at a general point ( a, exp(a) ) is now: f(x) = exp(a) * x + exp(a) * (1-a) = exp(a) * (x + 1 - a) .
Now for the second part of the question, simply put in ( a,exp(a) ) = (0, 1) into our above equation. This gives: f(x) = 1 * (x + 1 - 0) => f(x) = x + 1 as our final answer.