### Video Transcript

Find the third derivative of π¦ with respect to π₯ given that π¦ is equal to five over eight multiplied by the natural logarithm of four π₯.

In this question, weβre given a function π¦ in terms of π₯ and weβre asked to find the third derivative of π¦ with respect to π₯. So weβre going to need to differentiate this three times with respect to π₯. Letβs start by finding the first derivative of π¦ with respect to π₯. Thatβs the derivative of five over eight times the natural logarithm of four π₯ with respect to π₯.

To do this, we recall the derivative of the natural logarithm of ππ₯ with respect to π₯ is equal to one over π₯. And we can use this to differentiate this function by first taking the constant factor of five over eight outside of the derivative. We get five over eight multiplied by one over π₯ which simplifies to give us five over eight π₯. We can now find the second derivative of π¦ with respect to π₯ by differentiating five over eight π₯ with respect to π₯.

And weβre going to want to do this by using the power rule for differentiation. First, weβre going to rewrite five divided by eight π₯ as five over eight multiplied by π₯ to the power of negative one. This then allows us to differentiate this by using the power rule for differentiation which tells us for any real constants π and π the derivative of π times π₯ to the πth power with respect to π₯ is equal to π times π multiplied by π₯ to the power of π minus one. We multiply it by the exponent of π₯ and reduce this exponent by one.

In this case, our exponent of π₯ is negative one. So we multiply it by the exponent of π₯ which is negative one and then reduce this exponent by one. This gives us negative five over eight multiplied by π₯ to the power of negative two. Now weβre ready to find the third derivative of π¦ with respect to π₯. To do this, we need to differentiate the second derivative of π¦ with respect to π₯ with respect to π₯. So we need to differentiate negative five over eight times π₯ to the power of negative two with respect to π₯.

Once again, we can do this by using the power rule for differentiation. We multiply it by the exponent of π₯ and then reduce this exponent by one. We get negative two times negative five over eight multiplied by π₯ to the power of negative three. And we can simplify this. We have a negative multiplied by a negative, which is a positive and we can cancel the shared factor of two in the numerator and denominator. This gives us five over four times π₯ to the power of negative three, which we can then rewrite by using our laws of exponents. π₯ to the power of negative π is equal to one divided by π₯ to the πth power. This gives us five divided by four π₯ cubed, which is our final answer.

Therefore, we were able to show if π¦ is five divided by eight multiplied by the natural logarithm of four π₯, then the third derivative of π¦ with respect to π₯ is equal to five divided by four π₯ cubed.