Video: Differentiating Trigonometric Functions Using the Product Rule

If 𝑦 = 8π‘₯ cos 6π‘₯, find d𝑦/dπ‘₯.

02:43

Video Transcript

If 𝑦 is equal to eight π‘₯ times the cos of six π‘₯, find the derivative of 𝑦 with respect to π‘₯.

We need to find an expression for the derivative of 𝑦 with respect to π‘₯, and we can see that we’re given 𝑦 as the product of two functions. It’s the product of eight π‘₯ and the cos of six π‘₯, so we can try and find d𝑦 by dπ‘₯ by using the product’s rule. We recall the product rule tells us if 𝑦 is the product of two differentiable functions, 𝑒 of π‘₯ times 𝑣 of π‘₯, then the derivative of 𝑦 with respect to π‘₯ is equal to 𝑒 of π‘₯ times the derivative of 𝑣 with respect to π‘₯ plus 𝑣 of π‘₯ times the derivative of 𝑒 with respect to π‘₯.

In our case, 𝑦 is equal to eight π‘₯ multiplied by the cos of six π‘₯, so we’ll set 𝑒 of π‘₯ to be eight π‘₯ and 𝑣 of π‘₯ to be the cos of six π‘₯. And we know how to differentiate both of these expressions, so we can do this by using the product rule. To use the product rule, we need expressions for d𝑒 by dπ‘₯ and d𝑣 by dπ‘₯, so let’s calculate these first. Let’s start with d𝑒 by dπ‘₯. That’s the derivative of eight π‘₯ with respect to π‘₯. We could do this by using the power of the differentiation. We would rewrite eight π‘₯ as eight π‘₯ to the first power and then multiply by our exponent of π‘₯ and then reduce this exponent by one.

However, we can also do this by noticing eight π‘₯ is a linear function. And we know the slope of a linear function is just the coefficient of π‘₯, which in this case is eight. So we have d𝑒 by dπ‘₯ is equal to eight. We also need to find an expression for d𝑣 by dπ‘₯. That’s the derivative of the cos of six π‘₯ with respect to π‘₯. And to differentiate this expression, we need to recall one of our standard rules for differentiating trigonometric functions. We know for any real constant π‘Ž, the derivative of the cos of π‘Žπ‘₯ with respect to π‘₯ is equal to negative π‘Ž times the sin of π‘Žπ‘₯. In this case, our value of π‘Ž is equal to six, so we get the derivative of the cos of six π‘₯ with respect to π‘₯ is equal to negative six times the sin of six π‘₯.

Now that we’ve found expressions for d𝑒 by dπ‘₯ and d𝑣 by dπ‘₯, we can use the product rule to help us find d𝑦 by dπ‘₯. It’s equal to 𝑒 of π‘₯ times d𝑣 by dπ‘₯ plus 𝑣 of π‘₯ times d𝑒 by dπ‘₯. Substituting in our expressions for 𝑒 of π‘₯, 𝑣 of π‘₯, d𝑣 by dπ‘₯, and d𝑒 by dπ‘₯, we get that d𝑦 by dπ‘₯ is equal to eight π‘₯ times negative six sin of six π‘₯ plus the cos of six π‘₯ all multiplied by eight. And if we evaluate and rearrange this equation, we get negative 48π‘₯ times the sin of six π‘₯ plus eight times the cos of six π‘₯. And this is our final answer.

Therefore, we were able to show if 𝑦 is equal to eight π‘₯ times the cos of six π‘₯, then the derivative of 𝑦 with respect to π‘₯ is equal to negative 48π‘₯ times the sin of six π‘₯ plus eight multiplied by the cos of six π‘₯.

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