Question Video: Differentiate the Composition of a Trigonometric Function and a Logarithmic Function | Nagwa Question Video: Differentiate the Composition of a Trigonometric Function and a Logarithmic Function | Nagwa

# Question Video: Differentiate the Composition of a Trigonometric Function and a Logarithmic Function Mathematics • Third Year of Secondary School

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Differentiate the function π¦ = tan [ln (ππ₯ + π)].

03:10

### Video Transcript

Differentiate the function π¦ is equal to the tan of the natural logarithm of π π₯ plus π.

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 π¦ is a function of π’ and π’ is a function of π₯, then the derivative of π¦ with respect to π₯ is equal to the derivative of π¦ with respect to π’ multiplied by the derivative of π’ with respect to π₯. So, if we set our function π’ to be the natural logarithm of ππ₯ plus π, we see that π¦ is equal to the tangent of π’. Itβs now a function of π’. And π’ is a function of π₯. So, we can apply the chain rule.

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

Weβll start by setting π£ equal to the inner linear function ππ₯ plus π. This gives us that π’ is equal to the natural logarithm of π£. π’ is a function of π£. And π£ is a function of π₯. This means we can calculate the derivative of π’ with respect to π₯ by using the chain rule. The derivative of π’ with respect to π₯ is equal to the derivative of π’ with respect to π£ multiplied by the derivative of π£ with respect to π₯.

Substituting this into our expression for the derivative of π¦ with respect to π₯ gives us that dπ¦ dπ₯ is equal to dπ¦ by dπ’ times dπ’ by dπ£ times dπ£ by dπ₯. Weβre now ready to start evaluating this expression. Weβll start by dπ¦ by dπ’, which is the derivative of the tan of π’ with respect to π’. We know the derivative of the tan of π with respect to π is equal to the sec squared of π. So, the derivative of the tan of π’ with respect to π’ is the sec squared of π’.

Now, we need to calculate dπ’ by dπ£. Thatβs the derivative of the natural logarithm of π£ with respect to π£. 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 π£ with respect to π£ is one over π£. Finally, we need to find dπ£ by dπ₯. Thatβs the derivative of π₯π plus π with respect to π₯. The derivative of ππ₯ is just π. And the derivative of a constant π is zero. So, this evaluates to give us π. Multiplying all of these together gives us π times the sec squared of π’ all divided by π£.

Finally, we want to rewrite our answer in terms of π₯. So, we use that π’ is equal to the natural logarithm of ππ₯ plus π and π£ is equal to ππ₯ plus π to rewrite our answer as π times the sec squared of the natural logarithm of ππ₯ plus π all divided by ππ₯ plus π. Therefore, weβve shown the derivative of the function π¦ is equal to the tan of the natural logarithm of ππ₯ plus π is equal to π times the sec squared of the natural logarithm of ππ₯ plus π all divided by ππ₯ plus π.

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