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Question: How do you find the equation of the tangent line to the graph \(y = x{e^x} - {e^x}\) through point \...

How do you find the equation of the tangent line to the graph y=xex−exy = x{e^x} - {e^x} through point (1,0)\left( {1,0} \right)?

Explanation

Solution

We are given an equation of graph and we have to find the equation of tangent line to the graph and also given the graph through which it passes. To find the equation of the line we need a point and a slope. The slope of the tangent line is the value of the derivative at the point of tangency. To find the derivatives we will use the product rule which states that
F(x) = g(x).h(x)
F’(x) = g’(x)h(x) + g(x)h’(x)
Here g(x) and h(x) are two different functions. F’(x) , g’(x), h’(x) arte their derivatives. We know that the derivative of a function is equal to the rate of change of that of function at the point. Derivative at the point will be equal to the slope of the tangent line. After finding the slope of the line we will use the point slope form to find the equation of the tangent.
Y−y1=m(X−x1).Y-y_1=m(X-x_1).
Here y1y_1 and x1x_1 are the X and Y coordinates of the point through which line passes. Like this we will get the equation.

Complete step-by-step answer:
Step1: We are given an equation of graph i.e.y=xex−exy = x{e^x} - {e^x} and we have to find the equation of tangent to it. So to find the equation of tangent first we will find the slope which can be found by the derivative of the function.
For the function
y=f(x)=xex−exy = f(x) = x{e^x} - {e^x}
Factor outex{e^x}
⇒f(x)=ex(x−1)\Rightarrow f(x) = {e^x}(x - 1)
Apply the product rule to differentiate:
⇒f′(x)=ex(x−1)+ex(1)\Rightarrow f'(x) = {e^x}(x - 1) + {e^x}\left( 1 \right)
Simplifying the expression
f′(x)=ex(x−1+1)f'(x) = {e^x}\left( {x - 1 + 1} \right)
=xex= x{e^x}
Step2: Recall that the derivative of a function at a point is equal to the rate of change of that function at that point. In other words, the derivative of f(x) at(1,0)\left( {1,0} \right). Since we have found the derivative function of f(x), plug in 1 to f’(x) to get the slope of the tangent line.
⇒f′(1)=(1)e1=e\Rightarrow f'(1) = (1){e^1} = e
Now we have the slope of a line and one point that it goes through. We can apply the point-slope form of a line to find its equation.
Step3: point slope form is:
Y−y1=m(X−x1)Y-y_1=m(X-x_1)
Where y1y_1 is f(1), m=f,(1), and x1x_1 =1 = 1
Substituting the value in the equation we will get:
⇒y−0=e(x−1)\Rightarrow y - 0 = e(x - 1)
⇒y=xe−e\Rightarrow y = xe - e

Hence the equation of tangent is y=xe−ey = xe - e

Note:
In this question the student didn't get an approach how to solve the question. In order to find the equation of tangent of the line so first we have to find the slope and slope means derivative. Then simply use the point- slope form to find that in many cases slope is already given in the question so directly apply the point-slope form and use the given slope. This is the only method to find the equation of tangent
Commit to memory:
Y−y1=m(X−x1)Y-y_1=m(X-x_1)