Question Video: Finding the First Derivative of a Function Having a Variable in Its Base and Exponent Using Logarithmic Differentiation | Nagwa Question Video: Finding the First Derivative of a Function Having a Variable in Its Base and Exponent Using Logarithmic Differentiation | Nagwa

# Question Video: Finding the First Derivative of a Function Having a Variable in Its Base and Exponent Using Logarithmic Differentiation Mathematics • Third Year of Secondary School

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Find dπ¦/dπ₯ if 6π¦ = 7π₯^(3/5π₯).

05:22

### Video Transcript

Find dπ¦ by dπ₯ if six π¦ is equal to seven π₯ raised to the power three over five π₯.

Weβre asked to find the derivative with respect to π₯ of the function six π¦ is equal to seven π₯ raised to the power of three over five π₯. And before we begin, there are a couple of things we need to note about our equation here. The first is that on the left-hand side, we have six π¦ as opposed to π¦. Since π¦ is not by itself on the left, we call this an implicit function or equation. The second thing to notice that our exponent on the right-hand side is not a constant. In fact, our exponent involves a variable π₯. And because of this, we canβt use any of our usual rules for differentiation. What we can do to find dπ¦ by dπ₯ is to use logarithmic differentiation.

The first thing we can do is to isolate π¦ on the left-hand side by dividing through by six. We can then cancel the sixes on the left so that we have π¦ is equal to seven over six π₯ raised to the power three over five π₯. We now apply logarithmic differentiation. And our first step is to apply the natural logarithm to both sides. And remember that the natural logarithm is the logarithm to the base π, where, to five decimal places, π which is Eulerβs number is 2.71828 so that we have the natural logarithm of π¦ is the natural logarithm of π of π₯. In our case, thatβs the natural logarithm of seven over six times π₯ raised to the power three over five π₯.

We should note that for the logarithmic differentiation to be valid, we need to specify here that π¦ is greater than zero. Thatβs because the logarithm of zero is undefined and the function doesnβt exist for negative values. If we want to include negative values, we must include absolute value lines around π¦ and around π of π₯. But in our case, weβll simply assume that π¦ is greater than zero. Now, our function actually looks more complicated than it did in the first place. But this is where the use of the logarithms comes in. We use the laws of logarithms to expand our right-hand side. The first thing we can do is use the product rule for logs, which says that the log of ππ is equal to the log of π plus the log of π, so that on our right-hand side, we can split this up into the log of seven over six plus the natural log of π₯ raised to the power three over five π₯.

And we can expand further our second term using the power rule for logs. Thatβs the log of π raised to the power π is π times the log of π. Our right-hand side is then the natural logarithm of seven over six plus three over five π₯ times the natural logarithm of π₯. Our third step in logarithmic differentiation is to differentiate both sides with respect to π₯. On our left-hand side, weβre differentiating a function of a function because π¦ is actually a function of π₯. And remember that if we have a function π which is a function of π¦ which is a function of π₯, by the chain rule, we have dπ by dπ₯ is equal to dπ by dπ¦ times dπ¦ by dπ₯. And in our case, π is the natural logarithm of π¦.

And now, we can use the result that the derivative of the natural logarithm of π’ with respect to π’ is one over π’ for π’ greater than zero so that on our left-hand side, we have one over π¦ times dπ¦ by dπ₯, where one over π¦ is our dπ by dπ¦. And now, on the right-hand side, we know that the logarithm of seven over six is the logarithm of a constant is a constant, so the derivative of this is equal to zero. So now, we need to find the derivative of three over five π₯ times the logarithm of π₯. And for this, we can use the product rule for differentiation, where, in our case, π is three over five π₯ and π is the natural logarithm of π₯.

Noting that three over five π₯ is actually three over five times π₯ raised to the power negative one and this is a function of the form π times π₯ raised to the πth power, then we can use the power rule for derivatives, which says that d by dπ₯ of π times π₯ raised to the πth power is π times π times π₯ raised to the π minus oneth power. That is, we multiply by the exponent and subtract one from the exponent. d by dπ₯ is three over five π₯ raised to the power negative one is negative three over five π₯ raised to the power negative two.

And recalling that the derivative of the natural logarithm of π₯ is one over π₯, so that in our product rule π prime is negative three over five π₯ raised to the power negative two and π prime is one over π₯, the derivative on our right-hand side is negative three over five π₯ squared, which is π prime, times the natural logarithm of π₯, which is π, plus three over five π₯, which is π, times one over π₯, which is π prime. Rearranging this, we have three over five π₯ squared minus three over five π₯ squared times the natural logarithm of π₯. And since we have a common factor of three over five π₯ squared, we can take this outside.

Now, remember, weβre trying to find dπ¦ by dπ₯. This means we need to isolate dπ¦ by dπ₯ on the left-hand side. We can do this by multiplying through by π¦. And this is our fourth step in our logarithmic differentiation. We can cancel the π¦βs on the left-hand side. And on the right-hand side, we have three π¦ over five π₯ squared times one minus the logarithm of π₯. Remember, though, that π¦ is equal to seven over six times π₯ raised to the power three over five π₯. And substituting this in, we can cancel a three so that our constant is actually seven over 10. So now, if we look at our powers of π₯, this is actually π₯ raised to the power three over five π₯ minus two. And we have dπ¦ by dπ₯ is equal to seven over 10 times π₯ raised to the power three over five π₯ minus two times one minus the natural logarithm of π₯.

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