### Video Transcript

If ln π₯ plus π₯ multiplied by π to the power of π¦ equals one, find dπ¦ by dπ‘ when dπ₯ by dπ‘ equals five, π₯ equals one, and π¦ equals zero.

We could differentiate the equation in terms of π₯. Then, we could use the chain rule which states that dπ¦ by dπ₯ is equal to dπ¦ by dπ‘ multiplied by dπ‘ by dπ₯. However, in this case, it will be much easier to differentiate the equation in terms of π‘. We need to differentiate each of the three terms: ln π₯, π₯ multiplied by to π the power of π¦, and one in terms of π‘.

Differentiating ln π₯ with respect to π‘ gives us one over π₯ multiplied by dπ₯ dπ‘. In order to differentiate π₯ multiplied by π to the power of π¦, we must use the product rule. The product rule states that if π¦ is equal to π’ multiplied by π£ then dπ¦ by dπ₯, the differential, is equal to π’ multiplied by π£ dash plus π£ multiplied by π’ dash, where π’ dash and π£ dash are the differentials of π’ and π£, respectively.

In our question, we will let π’ equal π₯ and π£ equal π to the power of π¦. Remember weβre differentiating these expressions with respect to π‘. This means that π’ dash will be one multiplied by dπ₯ dπ‘. This is equivalent to dπ₯ dπ‘. Differentiating π to the power of π¦ with respect to π‘ is equal to π to the power of π¦ multiplied by dπ¦ dπ‘. Multiplying π’ and π£ dash gives us π₯π to the power of π¦ multiplied by dπ¦ dπ‘. Multiplying π£ and π’ dash gives us π to the power of π¦ multiplied by dπ₯ dπ‘. Finally, differentiating one with respect to π‘ gives us zero. Any constant differentiated is equal to zero.

We can now substitute in our values for dπ₯ dπ‘, π₯, and π¦. The first term becomes one over one multiplied by five. One divided by one is equal to one and one multiplied by five is equal to five. Therefore, one over π₯ multiplied by dπ₯ dπ‘ equals five. The second term becomes one multiplied by π to the power of zero multiplied by dπ¦ dπ‘. π to the power of zero is equal to one. And multiplying this by one also gives an answer of one. This means that the second term just simplifies to dπ¦ by dπ‘.

Substituting π¦ equals zero and dπ₯ by dπ‘ equals five into the third term gives us π to the power of zero multiplied by five. Once again, π to the power of zero is equal to one and multiplying one by five gives us five. We are therefore left with five plus dπ¦ by dπ‘ plus five equals zero. Collecting like terms by adding the two fives gives us dπ¦ by dπ‘ plus 10 is equal to zero. We can subtract 10 from both sides of this equation, giving us a final answer of dπ¦ by dπ‘ equals negative 10.

If ln π₯ plus π₯ multiplied by π to the power of π¦ is equal to one, then when dπ₯ by dπ‘ equals five, π₯ equals one, and π¦ equals zero, dπ¦ by dπ‘ will be equal to negative 10.