# Video: APCALC02AB-P1A-Q07-893181043695

What is the slope of the tangent to the curve π¦ = (π^(2π₯))/(1 + π₯) at π₯ = 2?

03:08

### Video Transcript

What is the slope of the tangent to the curve π¦ equals π to the power of two π₯ over one plus π₯ at π₯ equals two?

The slope of the tangent to the curve at a particular point is the same as the slope of the curve itself at that point, which we can find using differentiation. We first need to find the gradient function dπ¦ by dπ₯. In this case, we note first of all that π¦ is actually a quotient. Itβs equal to π to the power of two π₯ over one plus π₯, both of which are differentiable functions. So weβre going to need to apply the quotient rule in order to find dπ¦ by dπ₯.

The quotient rule tells us that if π¦ is equal to π’ over π£, where both π’ and π£ are differentiable functions of π₯, then dπ¦ by dπ₯ is equal to π£ times dπ’ by dπ₯ minus π’ times dπ£ by dπ₯ all over π£ squared. We therefore define π’ to be the numerator of the quotient thatβs π to the power of two π₯ and π£ to be the denominator thatβs one plus π₯. We then find the derivatives of both of π’ and π£ with respect to π₯. In order to find dπ’ by dπ₯, we must recall how to differentiate exponential functions.

We recall that the derivative with respect to π₯ of π to the power of ππ₯, where π is some constant, is ππ to the power of ππ₯. And so, the derivative of π to the power of two π₯ is two π to the two π₯. The derivative of one plus π₯ with respect to π₯ is just one. So we have both dπ’ by dπ₯ and dπ£ by dπ₯. Now, we can substitute into our formula for the quotient rule. dπ¦ by dπ₯ is equal to π£ times dπ’ by dπ₯βthatβs one plus π₯ times two π to the two π₯βminus π’ times dπ£ by dπ₯βthatβs π to power of two π₯ times one. But of course, multiplying by one has no effect. This is all over π£ squared. Thatβs one plus π₯ all squared.

Distributing the parentheses in the numerator, we see that dπ¦ by dπ₯ is equal to two π to the two π₯ plus two π₯π to the two π₯ minus π to the two π₯ over one plus π₯ squared. And then subtracting π to the two π₯ from two π to the two π₯, this simplifies to π to the power of two π₯ plus two π₯ π to the power of two π₯ all over one plus π₯ all squared. Now, this is the general gradient or slope function of our curve. In order to find the slope of the curve or the tangent to the curve at a given point, we need to evaluate dπ¦ by dπ₯ at that point. In this case, we need to evaluate dπ¦ by dπ₯ when π₯ is equal to two.

Substituting π₯ equals two then gives π to the power of two times two plus two times two π to the power of two times two over one plus two all squared. Simplifying gives π to the fourth power plus four π to the fourth power over three squared which is equal to nine. And, finally, combining the terms in the numerator gives five π to the fourth power over nine.

So by differentiating the equation of the curve and then evaluating this gradient or slope function dπ¦ by dπ₯ when π₯ equals two, we found that the slope of the tangent to the given curve at π₯ equals two is five π to the fourth power over nine.