Video: Finding the First Derivative of a Combination of Logarithmic Functions Using the Product and the Chain Rules

Find d𝑦/dπ‘₯, given that 𝑦 = 3π‘₯⁢ ln 3π‘₯Β³.

04:06

Video Transcript

Find d𝑦 dπ‘₯, given that 𝑦 is equal to three π‘₯ to the power of six ln three π‘₯ cubed.

So if we’re looking to differentiate this function, the first thing we can do is take the constant value, which is our three at the beginning, outside and then deal that afterwards because that won’t affect the differentiation. So therefore, we can say the derivative of 𝑦 is equal to three multiplied by the derivative of π‘₯ to the power of six ln three π‘₯ to the power of three. So what we can see now is that we’ve got π‘₯ to the power of six multiplied by ln three π‘₯ to the power of three. So we can use the product rule because it’s like a function multiplied by another function.

And the product rule tells us that if we have 𝑓 of π‘₯ multiplied by 𝑔 of π‘₯, then if we say that 𝑒 is equal to 𝑓 of π‘₯ and 𝑣 is equal to 𝑔 of π‘₯, then derivative of 𝑒𝑣 is equal to 𝑒 d𝑣 dπ‘₯ plus 𝑣 d𝑒 dπ‘₯ which is our first term multiplied by the derivative of the second term plus our second term multiplied by the derivative of the first term. Okay great, now, we have that; we can apply that to solve our problem.

So first of all, we can say the 𝑒 is equal to π‘₯ to the power of six. And therefore if we differentiate this, we’re gonna get six π‘₯ to the power of five. That’s because what we do is we multiply the exponent by the coefficient, so six by one which gives us six. And then we reduce the exponent by one. So we go from six to five. So we got six π‘₯ to the power of five. And then next, we can say that 𝑣 is equal to ln three π‘₯ cubed.

And to be able to differentiate this, what we can use is an adaptation of the chain rule. And this tells us that the derivative of ln 𝑒 of π‘₯ is equal to one over 𝑒 of π‘₯ multiplied by the derivative of 𝑒 of π‘₯. And in our expression, the three π‘₯ cubed is our 𝑒 of π‘₯. So therefore, we can say that d𝑣 dπ‘₯ is gonna be equal to one over three π‘₯ cubed multiplied by ⁠— then we’ve got nine π‘₯ squared, and that’s because this is the derivative of three π‘₯ cubed. So again, exponent multiplied by coefficient is three multiplied by three which is nine. Then we reduce the exponent by one to two.

Okay great, we can now simplify this. And to first simplify, we’re gonna divide the top by three and the bottom by three. So we’re gonna get three π‘₯ squared. And we’re gonna have π‘₯ cubed on the bottom. And then we can divide three by π‘₯ squared. So then we’re left with one multiplied by three on the numerator and just π‘₯ in the denominator. So we’re left with d𝑣 dπ‘₯ is equal to three over π‘₯. So now, it’s time to apply our product rule and bring everything back together. So first fist of all, we’re gonna have three multiplied by β€” then we’re gonna have three π‘₯ to the power of six over π‘₯. And that’s because that’s our 𝑒d𝑣 dπ‘₯ because 𝑒 was π‘₯ to the power of six. And d𝑣 dπ‘₯ was three over π‘₯. And this is gonna be plus and then three π‘₯ cubed multiplied by six π‘₯ to the power of five. And this is because this is our 𝑣d𝑒 dπ‘₯.

Well now, we can look to simplify. So first of all, in the first term, what we can do is divide the top and bottom or numerator and denominator by π‘₯. So we’re left with three π‘₯ to the power of five. So then what we can do is we can distribute the three across the parentheses. And when we do this we get nine π‘₯ to the power of five plus 18π‘₯ to the power of five ln three π‘₯ cubed. So therefore, we can say that, given that 𝑦 is equal to three π‘₯ to the power of six ln three π‘₯ cubed, then d𝑦 dπ‘₯ is equal to nine π‘₯ to the power of five plus 18π‘₯ to the power of five ln three π‘₯ cubed.

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