Video: APCALC02AB-P1A-Q07-893181043695

What is the slope of the tangent to the curve 𝑦 = (𝑒^(2π‘₯))/(1 + π‘₯) at π‘₯ = 2?

03:08

Video Transcript

What is the slope of the tangent to the curve 𝑦 equals 𝑒 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 d𝑦 by dπ‘₯. In this case, we note first of all that 𝑦 is actually a quotient. It’s equal to 𝑒 to the power of two π‘₯ over one plus π‘₯, both of which are differentiable functions. So we’re going to need to apply the quotient rule in order to find d𝑦 by dπ‘₯.

The quotient rule tells us that if 𝑦 is equal to 𝑒 over 𝑣, where both 𝑒 and 𝑣 are differentiable functions of π‘₯, then d𝑦 by dπ‘₯ is equal to 𝑣 times d𝑒 by dπ‘₯ minus 𝑒 times d𝑣 by dπ‘₯ all over 𝑣 squared. We therefore define 𝑒 to be the numerator of the quotient that’s 𝑒 to the power of two π‘₯ and 𝑣 to be the denominator that’s one plus π‘₯. We then find the derivatives of both of 𝑒 and 𝑣 with respect to π‘₯. In order to find d𝑒 by dπ‘₯, we must recall how to differentiate exponential functions.

We recall that the derivative with respect to π‘₯ of 𝑒 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 𝑒 to the power of two π‘₯ is two 𝑒 to the two π‘₯. The derivative of one plus π‘₯ with respect to π‘₯ is just one. So we have both d𝑒 by dπ‘₯ and d𝑣 by dπ‘₯. Now, we can substitute into our formula for the quotient rule. d𝑦 by dπ‘₯ is equal to 𝑣 times d𝑒 by dπ‘₯β€”that’s one plus π‘₯ times two 𝑒 to the two π‘₯β€”minus 𝑒 times d𝑣 by dπ‘₯β€”that’s 𝑒 to power of two π‘₯ times one. But of course, multiplying by one has no effect. This is all over 𝑣 squared. That’s one plus π‘₯ all squared.

Distributing the parentheses in the numerator, we see that d𝑦 by dπ‘₯ is equal to two 𝑒 to the two π‘₯ plus two π‘₯𝑒 to the two π‘₯ minus 𝑒 to the two π‘₯ over one plus π‘₯ squared. And then subtracting 𝑒 to the two π‘₯ from two 𝑒 to the two π‘₯, this simplifies to 𝑒 to the power of two π‘₯ plus two π‘₯ 𝑒 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 d𝑦 by dπ‘₯ at that point. In this case, we need to evaluate d𝑦 by dπ‘₯ when π‘₯ is equal to two.

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

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

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.