Video: EG19M1-DiffAndInt-Q11

EG19M1-DiffAndInt-Q11

04:19

Video Transcript

If ln π‘₯ plus π‘₯ multiplied by 𝑒 to the power of 𝑦 equals one, find d𝑦 by d𝑑 when dπ‘₯ by d𝑑 equals five, π‘₯ equals one, and 𝑦 equals zero.

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

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

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

We can now substitute in our values for dπ‘₯ d𝑑, π‘₯, and 𝑦. 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‘₯ d𝑑 equals five. The second term becomes one multiplied by 𝑒 to the power of zero multiplied by d𝑦 d𝑑. 𝑒 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 d𝑦 by d𝑑.

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

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

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