# Video: EG19M1-DiffAndInt-Q11

EG19M1-DiffAndInt-Q11

04:19

### Video Transcript

If ln 𝑥 plus 𝑥 multiplied by 𝑒 to the power of 𝑦 equals one, find d𝑦 by d𝑡 when d𝑥 by d𝑡 equals five, 𝑥 equals one, and 𝑦 equals zero.

We could differentiate the equation in terms of 𝑥. Then, we could use the chain rule which states that d𝑦 by d𝑥 is equal to d𝑦 by d𝑡 multiplied by d𝑡 by d𝑥. However, in this case, it will be much easier to differentiate the equation in terms of 𝑡. We need to differentiate each of the three terms: ln 𝑥, 𝑥 multiplied by to 𝑒 the power of 𝑦, and one in terms of 𝑡.

Differentiating ln 𝑥 with respect to 𝑡 gives us one over 𝑥 multiplied by d𝑥 d𝑡. In order to differentiate 𝑥 multiplied by 𝑒 to the power of 𝑦, we must use the product rule. The product rule states that if 𝑦 is equal to 𝑢 multiplied by 𝑣 then d𝑦 by d𝑥, the differential, is equal to 𝑢 multiplied by 𝑣 dash plus 𝑣 multiplied by 𝑢 dash, where 𝑢 dash and 𝑣 dash are the differentials of 𝑢 and 𝑣, respectively.

In our question, we will let 𝑢 equal 𝑥 and 𝑣 equal 𝑒 to the power of 𝑦. Remember we’re differentiating these expressions with respect to 𝑡. This means that 𝑢 dash will be one multiplied by d𝑥 d𝑡. This is equivalent to d𝑥 d𝑡. Differentiating 𝑒 to the power of 𝑦 with respect to 𝑡 is equal to 𝑒 to the power of 𝑦 multiplied by d𝑦 d𝑡. Multiplying 𝑢 and 𝑣 dash gives us 𝑥𝑒 to the power of 𝑦 multiplied by d𝑦 d𝑡. Multiplying 𝑣 and 𝑢 dash gives us 𝑒 to the power of 𝑦 multiplied by d𝑥 d𝑡. Finally, differentiating one with respect to 𝑡 gives us zero. Any constant differentiated is equal to zero.

We can now substitute in our values for d𝑥 d𝑡, 𝑥, 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 𝑥 multiplied by d𝑥 d𝑡 equals five. The second term becomes one multiplied by 𝑒 to the power of zero multiplied by d𝑦 d𝑡. 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 d𝑡.

Substituting 𝑦 equals zero and d𝑥 by d𝑡 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 d𝑡 plus five equals zero. Collecting like terms by adding the two fives gives us d𝑦 by d𝑡 plus 10 is equal to zero. We can subtract 10 from both sides of this equation, giving us a final answer of d𝑦 by d𝑡 equals negative 10.

If ln 𝑥 plus 𝑥 multiplied by 𝑒 to the power of 𝑦 is equal to one, then when d𝑥 by d𝑡 equals five, 𝑥 equals one, and 𝑦 equals zero, d𝑦 by d𝑡 will be equal to negative 10.