# Video: Finding the Slope of the Tangent to a Graph Using Differentiation of Trigonometric Functions

Suppose that the function π satisfies π(π₯) = 2 cos (ππ₯) β 4π₯Β³. What is the slope of the line tangent to the graph of π at the point π₯ = 1/2?

03:01

### Video Transcript

Suppose that the function π satisfies π of π₯ equals two cos ππ₯ minus four π₯ cubed. What is the slope of the line tangent to the graph of π at the point π₯ equals one-half?

If the line is tangent to the graph of a function at a particular point, then it has the same slope as the curve at that point. We know that we can find the slope of a curve by evaluating its first derivative at that point. We therefore need to find an expression for the first derivative, π prime of π₯. Now, looking at our function π of π₯, itβs composed of two terms. One is a trigonometric term, two cos of ππ₯. And one is a polynomial term, negative four π₯ cubed. So, we need to recall two rules of differentiation.

Firstly, for the trigonometric term, we recall that the derivative with respect to π₯ of cos of ππ₯ for some constant π is equal to negative π sin ππ₯. And remember, π is just a constant. The factor of two is a multiplicative constant, so we can just multiply our derivative by two. We have then that the derivative of the first term will be equal to two multiplied by negative π sin of ππ₯.

To differentiate the second term, we need to recall how to find the derivative of a polynomial term. We recall that the derivative with respect to π₯ of ππ₯ to the power of π for real values of π and π is equal to πππ₯ to the power of π minus one. We multiply by the exponent and then reduce the exponent by one. So, the derivative of our second term will be equal to negative four multiplied by three π₯ squared. Simplifying each term slightly and we have that π prime of π₯ is equal to negative two π sin of ππ₯ minus 12π₯ squared.

Now, this gives a general expression for the slope function of this curve. But we want to know the slope of the line tangent at a particular point, the point where π₯ is equal to one-half. So, we now need to evaluate this first derivative at π₯ equals one-half. Substituting π₯ equals one-half throughout our derivative gives that π prime of one-half is equal to negative two π sin of π by two minus 12 multiplied by one-half squared. We should know that sin of π by two is equal to one. Itβs one of those exact trigonometric values which we should know of by heart. Or we could also see this by sketching the graph of sin π₯.

One-half squared is equal to one-quarter. So, we have negative two π multiplied by one minus 12 multiplied by one-quarter. 12 multiplied by one-quarter is the same as 12 divided by four, which is three. So, we have our answer to the problem. By finding the first derivative of our function, π prime of π₯, and then evaluating it when π₯ equals a half, weβve found that the slope of the line tangent to the graph of π at this point is negative two π minus three.