Imagine being given the equation y=x³-2x+3, and being asked to find the tangent to the curve at the point where x=1.

The tangent to the curve will be a straight line, and therefore will take the form y=mx+c.

To find m (the gradient of the tangent), it is necessary first of all to differentiate the equation of the original curve. Doing this gives y’=3x²-2, where y’ is the gradient of the curve at a particular point. We are looking for the gradient at the point where x=1. Therefore, to find m, we must substitute x=1 into our expression. Doing so, we find that m=1.

We now know the equation of the tangent is y=x+c. To find c (the y-intercept), we must first of all know the coordinates of a point that the tangent is going to pass through. In our case, we know that the tangent must pass through the point on the line where x=1. To find the y-coordinate of this point, we can sub x=1 into our original equation of the curve. Doing so, we find that the point we must use is (1,2).

Now that we know a point on the line, we can sub those x and y values into the expression y=x+c. This gives us the equation 2=1+c, and some quick rearrangement shows us that c=1.

Therefore the equation of the tangent is y=x+1.

In summary:

-Equation of tangent is of the form y=mx+c

-To find m, differentiate the equation of the curve to find its gradient at the required point

-Find the coordinates of a point the tangent is going to pass through, and sub into the equation of the tangent to find c.

Differentiate y=x^3+ 7x-ln(2)

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