Video: Differentiate the Composition of a Trigonometric Function and a Logarithmic Function

Differentiate the function 𝑦 = tan [ln (π‘Žπ‘₯ + 𝑏)].

03:10

Video Transcript

Differentiate the function 𝑦 is equal to the tan of the natural logarithm of π‘Ž π‘₯ plus 𝑏.

The question is asking us to differentiate the composition of three functions. We have an innermost linear function, then we have a logarithmic function, and finally we have a trigonometric function. We can differentiate the composition of functions by using the chain rule.

The chain rule tells us that if 𝑦 is a function of 𝑒 and 𝑒 is a function of π‘₯, then the derivative of 𝑦 with respect to π‘₯ is equal to the derivative of 𝑦 with respect to 𝑒 multiplied by the derivative of 𝑒 with respect to π‘₯. So, if we set our function 𝑒 to be the natural logarithm of π‘Žπ‘₯ plus 𝑏, we see that 𝑦 is equal to the tangent of 𝑒. It’s now a function of 𝑒. And 𝑒 is a function of π‘₯. So, we can apply the chain rule.

The chain rule tells us the derivative of 𝑦 with respect to π‘₯ is equal to the derivative of 𝑦 with respect to 𝑒 multiplied by the derivative of 𝑒 with respect to π‘₯. At this point, we can calculate the derivative of 𝑦 with respect to 𝑒. However, the derivative of 𝑒 with respect to π‘₯ is itself the derivative of the composition of two functions. So, we’ll need to apply the chain rule again.

We’ll start by setting 𝑣 equal to the inner linear function π‘Žπ‘₯ plus 𝑏. This gives us that 𝑒 is equal to the natural logarithm of 𝑣. 𝑒 is a function of 𝑣. And 𝑣 is a function of π‘₯. This means we can calculate the derivative of 𝑒 with respect to π‘₯ by using the chain rule. The derivative of 𝑒 with respect to π‘₯ is equal to the derivative of 𝑒 with respect to 𝑣 multiplied by the derivative of 𝑣 with respect to π‘₯.

Substituting this into our expression for the derivative of 𝑦 with respect to π‘₯ gives us that d𝑦 dπ‘₯ is equal to d𝑦 by d𝑒 times d𝑒 by d𝑣 times d𝑣 by dπ‘₯. We’re now ready to start evaluating this expression. We’ll start by d𝑦 by d𝑒, which is the derivative of the tan of 𝑒 with respect to 𝑒. We know the derivative of the tan of πœƒ with respect to πœƒ is equal to the sec squared of πœƒ. So, the derivative of the tan of 𝑒 with respect to 𝑒 is the sec squared of 𝑒.

Now, we need to calculate d𝑒 by d𝑣. That’s the derivative of the natural logarithm of 𝑣 with respect to 𝑣. And we know the derivative of the natural algorithm of π‘₯ with respect to π‘₯ is equal to one over π‘₯. So, the derivative of the natural logarithm of 𝑣 with respect to 𝑣 is one over 𝑣. Finally, we need to find d𝑣 by dπ‘₯. That’s the derivative of π‘₯π‘Ž plus 𝑏 with respect to π‘₯. The derivative of π‘Žπ‘₯ is just π‘Ž. And the derivative of a constant 𝑏 is zero. So, this evaluates to give us π‘Ž. Multiplying all of these together gives us π‘Ž times the sec squared of 𝑒 all divided by 𝑣.

Finally, we want to rewrite our answer in terms of π‘₯. So, we use that 𝑒 is equal to the natural logarithm of π‘Žπ‘₯ plus 𝑏 and 𝑣 is equal to π‘Žπ‘₯ plus 𝑏 to rewrite our answer as π‘Ž times the sec squared of the natural logarithm of π‘Žπ‘₯ plus 𝑏 all divided by π‘Žπ‘₯ plus 𝑏. Therefore, we’ve shown the derivative of the function 𝑦 is equal to the tan of the natural logarithm of π‘Žπ‘₯ plus 𝑏 is equal to π‘Ž times the sec squared of the natural logarithm of π‘Žπ‘₯ plus 𝑏 all divided by π‘Žπ‘₯ plus 𝑏.

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