Video Transcript
Find the slope of the tangent to the curve π¦ is equal to negative eight log base five of seven π₯ squared plus one all squared at π₯ is equal to negative two rounded to the nearest hundredth.
In this question, weβre given the equation of a curve, and we can see that this is a very complicated curve. Weβre asked to find the slope of the tangent to this curve when π₯ is equal to negative two. And we need to round our answer to the nearest hundredth. To answer this question, we can start by recalling the slope of the tangent to a curve is given by its first derivative. In particular to find the slope of this curve when π₯ is negative two, we need to find the value of the first derivative of π¦ with respect to π₯ when π₯ is equal to negative two. Therefore, weβre going to need to differentiate our curve. However, doing this is very complicated because we can see that π¦ is equal to the composition of several functions.
So, we could differentiate this in a few different ways. Weβre going to start by simplifying the expression for our curve by using the laws of logarithms. We can recall the power rule for logarithms tells us for any logarithm base π, the log base π of π to the πth power is equal to π multiplied by the log base π of π. And we can use this to simplify our logarithmic expression since weβre taking the logarithm of an expression squared. Therefore, removing the exponent from the logarithm and multiplying the expression by two, we get that π¦ is equal to negative 16 times the logarithm base five of seven π₯ squared plus one.
We can now find an expression for dπ¦ by dπ₯. And to do this, we need to note that π¦ is the composition of two functions. So, weβre going to need to differentiate this by using the chain rule. And we can recall the chain rule tells us if π is a differentiable function at π₯ and π is a differentiable function at π of π₯, then the derivative of π composed with π of π₯ with respect to π₯ is equal to π prime of π₯ multiplied by π prime evaluated at π of π₯. We can apply this to our curve by setting π to be our inner function seven π₯ squared plus one and π to be the outer function negative 16 times the log base five of π. Doing this then gives us that π¦ is equal to π evaluated at π of π₯. And we can verify this by substituting the expression for π of π₯ into our function π of π.
This then allows us to evaluate dπ¦ by dπ₯ by using the chain rule. We need to find expressions for π prime and π prime. Letβs start with π prime. We need to differentiate negative 16 times the log base five of π with respect to π. And we can do this by recalling for any logarithm base π, the derivative of log base π of π₯ with respect to π₯ is equal to one divided by π₯ times the natural logarithm of π. In our case, our base is equal to five. And remember, multiplying by the constant value of negative 16 wonβt change the derivative. Applying this, we can show π prime of π is equal to negative 16 divided by π times the natural logarithm of five.
We now want to find an expression for π prime of π₯. Since π of π₯ is a polynomial, we can do this by using the power rule for differentiation. We apply this term by term. We multiply by the exponent of π₯ and reduce this exponent by one. The derivative of seven π₯ squared is 14π₯, and the derivative of the constant one is just equal to zero. We can now find an expression for dπ¦ by dπ₯ by substituting our expressions for π prime, π prime, and π into the chain rule. Doing this gives us dπ¦ by dπ₯ is equal to 14π₯ multiplied by negative 16 over seven π₯ squared plus one times the natural logarithm of five. Weβre not done yet, however. Remember, we need to use this to find the slope of the tangent to this curve when π₯ is equal to negative two. So, we need to evaluate this expression at π₯ is equal to negative two.
Substituting π₯ is equal to negative two into our expression for dπ¦ by dπ₯ gives us 14 times negative two multiplied by negative 16 divided by seven times negative two squared plus one multiplied by the natural logarithm of five. Evaluating this expression, we get 9.598 and this expansion continues. But remember, weβre asked to find this value to the nearest hundredth. Since the third decimal digit is eight, we need to round this value up. This gives us 9.60, which is our final answer.
Therefore, we were able to show the slope of the tangent to the curve π¦ is equal to negative eight times the log base five of seven π₯ squared plus one squared at π₯ is equal to negative two rounded to the nearest hundredth is 9.60.