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

Suppose that the function 𝑔 satisfies 𝑔(π‘₯) = 2 cos (πœ‹π‘₯) βˆ’ 4π‘₯Β³. What is the slope of the line tangent to the graph of 𝑔 at the point π‘₯ = 1/2?

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Video Transcript

Suppose that the function 𝑔 satisfies 𝑔 of π‘₯ equals two cos πœ‹π‘₯ minus four π‘₯ cubed. What is the slope of the line tangent to the graph of 𝑔 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, 𝑔 prime of π‘₯. Now, looking at our function 𝑔 of π‘₯, it’s composed of two terms. One is a trigonometric term, two cos of πœ‹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‘Žπ‘₯ for some constant π‘Ž is equal to negative π‘Ž sin π‘Žπ‘₯. And remember, πœ‹ 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œ‹ sin of πœ‹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‘Žπ‘₯ to the power of 𝑛 for real values of π‘Ž and 𝑛 is equal to π‘Žπ‘›π‘₯ to the power of 𝑛 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 𝑔 prime of π‘₯ is equal to negative two πœ‹ sin of πœ‹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 𝑔 prime of one-half is equal to negative two πœ‹ sin of πœ‹ by two minus 12 multiplied by one-half squared. We should know that sin of πœ‹ by two is equal to one. It’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œ‹ 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, 𝑔 prime of π‘₯, and then evaluating it when π‘₯ equals a half, we’ve found that the slope of the line tangent to the graph of 𝑔 at this point is negative two πœ‹ minus three.

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