Video Transcript
Differentiate 𝑔 of 𝑡 is equal to the square root of four times the natural logarithm of 𝑡 minus nine.
We’re given a function 𝑔 of 𝑡, and we’re asked to differentiate this function. To start, we notice that 𝑔 is a function in 𝑡. This means we need to differentiate this with respect to 𝑡. And to do this, we need to notice something about 𝑔 of 𝑡. It’s given as the composition of two functions. We’re taking the square root of four times the natural logarithm of 𝑡 minus nine.
And we know how to differentiate both of these functions. This means we could do this by using the chain rule. This would be a perfectly valid way of answering this question. However, because our outer function is the square root function, we could also do this by using the general power rule. To do this, we first need to remember one of our laws of exponents. The square root of 𝑎 is the same as 𝑎 to the power of one-half.
So by using this, we can rewrite 𝑔 of 𝑡 as four times the natural logarithm of 𝑡 minus nine all raised to the power of one-half. And now that we’ve rewritten 𝑔 in this form, we can differentiate it by using the general power rule. We recall the general power rule tells us for a differentiable function 𝑓 of 𝑡 and any constant 𝑛, the derivative of 𝑓 of 𝑡 all raised to the 𝑛th power with respect to 𝑡 is equal to 𝑛 times 𝑓 prime of 𝑡 multiplied by 𝑓 of 𝑡 raised to the power of 𝑛 minus one.
So in our case, we need to set our inner function four times the natural logarithm of 𝑡 minus nine to be 𝑓 of 𝑡 and our exponent 𝑛 equal to one-half. So we’re almost ready to find an expression for our derivative of 𝑔 of 𝑡 by using the general power rule. However, first, we need to find an expression for 𝑓 prime of 𝑡. That’s the derivative of four times the natural logarithm of 𝑡 minus nine with respect to 𝑡.
And we can just evaluate this derivative term by term. First, we need to remember the derivative of the natural logarithm of 𝑡 with respect to 𝑡 is equal to one over 𝑡. This tells us the derivative of four times the natural logarithm of 𝑡 with respect to 𝑡 is four over 𝑡. And of course the derivative of the constant negative nine is equal to zero.
So we’re now ready to find an expression for 𝑔 prime of 𝑡 by using the general power rule. We get it’s equal to 𝑛 times 𝑓 prime of 𝑡 multiplied by 𝑓 of 𝑡 all raised to the power of 𝑛 minus one. Substituting in 𝑛 is equal to one-half and our expressions for 𝑓 of 𝑡 and 𝑓 prime of 𝑡, we get that 𝑔 prime of 𝑡 is equal to one-half times four over 𝑡 multiplied by four times the natural logarithm of 𝑡 minus nine all raised to the power of one-half minus one.
And now we can start simplifying. First, our exponent of one-half minus one can be simplified to give us negative one-half. Next, we have one-half multiplied by four. We can simplify this to give us two. And we could leave our answer like this. However, there’s one more piece of simplification we’ll do.
Remember, by using our laws of exponents, 𝑎 to the power of negative one-half is the same as one divided by the square root of 𝑎. So we’ll use this to write our final factor in the denominator. This gives us two divided by 𝑡 times the square root of four times the natural logarithm of 𝑡 minus nine. And this is our final answer.
Therefore, we were able to show if 𝑔 of 𝑡 is equal to the square root of four times the natural logarithm of 𝑡 minus nine, then by using the general power rule or by using the chain rule, we can find that 𝑔 prime of 𝑡 is equal to two divided by 𝑡 times the square root of four times the natural logarithm of 𝑡 minus nine.
