Video Transcript
Find d𝑦 by d𝑥 given that seven 𝑦 is equal to six times 𝑥 raised to the power sin six 𝑥.
We’re given that seven 𝑦 is equal to six times 𝑥 raised to the power of sin six 𝑥. And we want to find the derivative of 𝑦 with respect to 𝑥. That’s d𝑦 by d𝑥. Now our right-hand side is quite complicated since we have the variable 𝑥 raised to the power sin six 𝑥, which is itself a function of 𝑥. And because of this, our usual methods of differentiation such as the chain product or quotient rules can’t be applied directly. What we can do, however, is use logarithmic differentiation.
And how this works is given 𝑦 is a function of 𝑥 is we first apply the natural logarithm to both sides so that we have the natural logarithm of 𝑦 is equal to the natural logarithm of 𝑓 of 𝑥, recalling that the natural logarithm is log to the base 𝑒, where 𝑒 is Euler’s number, which is approximately 2.71828 and so on. So let’s apply this to our function. But let’s first divide through by seven so we have 𝑦 on the left-hand side so that we have the natural logarithm of 𝑦 is equal to the natural logarithms of six over seven 𝑥 to the power of sin six 𝑥. And let’s just put parentheses around our argument to make it clear.
Now for this to be valid, we need to specify that 𝑦 is greater than zero. That’s because the log of zero is undefined and the log doesn’t exist for negative values. If we do want to include negative values, we need to use the absolute value signs for 𝑦 and 𝑓 of 𝑥 but 𝑦 is still nonzero. In our case, however, we’ll simply specify that 𝑦 is greater than zero. Now our second step in logarithmic differentiation is to use the laws of logarithms to simplify or expand.
In the first instance, we can use the product rule for logarithms on our right-hand side. This says that the logarithm to the base 𝑎 of the product 𝑏𝑐 is logarithm to the base 𝑎 of 𝑏 plus logarithm to the base 𝑎 of 𝑐. In our case, our base is 𝑒. Six over seven corresponds to 𝑏. And 𝑥 to power of sin six 𝑥 corresponds to 𝑐. Our product rule for logarithms then gives us the natural logarithm of 𝑦 is the natural logarithm of six over seven plus the natural logarithm of 𝑥 raised to the power sin six 𝑥.
Our second term on the right-hand side is still quite complicated. And to simplify this, we can use the power rule for logarithms. This says that log to the base 𝑎 of 𝑏 raised to the power 𝑐 is 𝑐 times log to the base 𝑎 of 𝑏. Our logarithm is to the base 𝑒 and now 𝑥 corresponds to 𝑏 and sin six 𝑥 corresponds to 𝑐. So now we bring our exponent sin six 𝑥 to the front and multiply that by the natural logarithm of 𝑥. And we have the natural logarithm of 𝑦 is equal to the natural logarithm of six over seven plus sin six 𝑥 times the natural logarithm of 𝑥.
Okay, so making some room, we have something manageable that we can easily differentiate. And this brings us to our third step in logarithmic differentiation. We differentiate both sides with respect to 𝑥. On our right-hand side, we can use the fact that the derivative of the sum is the sum of the derivatives. And the first thing to notice is that since the natural logarithm of six over seven is a constant, its derivative is equal to zero. On our left-hand side, we can use the fact that d by d𝑥 of the natural logarithm of 𝑦 is one over 𝑦 times d𝑦 by d𝑥, where 𝑦 is greater than zero for 𝑦 a function of 𝑥.
And for the final term on our right-hand side, we have the derivative of a product. And for this, we can use the product rule for differentiation. That is the derivative of a product, 𝑢𝑣, with respect to 𝑥 is 𝑢 times d𝑣 by d𝑥 plus d𝑢 by d𝑥 times 𝑣, where 𝑢 and 𝑣 are functions of 𝑥. If in our case we have 𝑢 is equal to sin six 𝑥 and 𝑣 is the natural logarithm of 𝑥, then we have d𝑢 by d𝑥 is six times the cos of six 𝑥 and d𝑣 by d𝑥 is one over 𝑥. Then by the product rule, the derivative with respect to 𝑥 of sin six 𝑥 times the natural logarithm of 𝑥 is sin six 𝑥, which is 𝑢, times one over 𝑥, which is d𝑣 by d𝑥, plus six times the cos of six 𝑥, which is d𝑢 by d𝑥, times the natural logarithm of 𝑥, which is 𝑣.
Okay, so making some room, now we can write this out more clearly. And we have one over 𝑦 d𝑦 by d𝑥 is equal to sin six 𝑥 over 𝑥 plus six cos six 𝑥 times the natural logarithm of 𝑥. We’re not quite finished yet, though, since it’s actually d𝑦 by d𝑥 that we want. So we’re going to need to eliminate the factor of one over 𝑦 on the left-hand side. And we can do this by multiplying both sides by 𝑦. On the left-hand side, this cancels. So we’re left with d𝑦 by d𝑥 on the left-hand side. And on the right-hand side, we recall that 𝑦 is actually six over seven times 𝑥 raised to the power sin six 𝑥.
And we’re now in step four of our logarithmic differentiation, which is solve for d𝑦 by d𝑥. So now substituting in 𝑦 is equal to six over seven 𝑥 raised to the power sin six 𝑥, we have d𝑦 by d𝑥 is six over seven times 𝑥 raised to the power sin six 𝑥 multiplied by sin of six 𝑥 all over 𝑥 plus six times the cos of six 𝑥 times the natural logarithm of 𝑥.
