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Question: How do you find the equation of the tangent line to the graph \[y = {x^2}{e^x} - 2x{e^x} + 2{e^x}\] ...

How do you find the equation of the tangent line to the graph y=x2ex−2xex+2exy = {x^2}{e^x} - 2x{e^x} + 2{e^x} through the point (1,e)\left( {1,e} \right)?

Explanation

Solution

Hint : Here in this question, we have to find the equation of the tangent line, given the x and y value, then find the slope of the equation i.e., m=dydxm = \dfrac{{dy}}{{dx}} at the point (x,y)\left( {x,y} \right), then find the equation of tangent line by using the equation y−y0=m(x−x0)y - {y_0} = m\left( {x - {x_0}} \right) where (x0,y0)\left( {{x_0},{y_0}} \right) is the point where tangent line passes through.

Complete step-by-step answer :
The tangent line to a curve at a given point is a straight line that just "touches" the curve at that point.
Consider the given equation which passes through the point (1,e)\left( {1,e} \right)
y=x2ex−2xex+2exy = {x^2}{e^x} - 2x{e^x} + 2{e^x}------(1)
Now, find the slope of the equation at the point of tangency (1,e)\left( {1,e} \right) using the first order derivative of given equation i.e., m=dydxm = \dfrac{{dy}}{{dx}}
⇒dydx=x2ex+2xex−2(xex+ex)+2ex\Rightarrow \dfrac{{dy}}{{dx}} = {x^2}{e^x} + 2x{e^x} - 2\left( {x{e^x} + {e^x}} \right) + 2{e^x}
⇒dydx=x2ex+2xex−2xex−2ex+2ex\Rightarrow \dfrac{{dy}}{{dx}} = {x^2}{e^x} + 2x{e^x} - 2x{e^x} - 2{e^x} + 2{e^x}
On simplification, we get
⇒m=dydx=x2ex\Rightarrow m = \dfrac{{dy}}{{dx}} = {x^2}{e^x}
The slope of the equation at the point (1,e)\left( {1,e} \right) is:
⇒(dydx)(1,e)=12e1\Rightarrow {\left( {\dfrac{{dy}}{{dx}}} \right)_{\left( {1,e} \right)}} = {1^2}{e^1}
On simplification, we get
⇒(dydx)(1,e)=e\Rightarrow {\left( {\dfrac{{dy}}{{dx}}} \right)_{\left( {1,e} \right)}} = e
The equation of slope of the tangent line is:
y−y0=m(x−x0)y - {y_0} = m\left( {x - {x_0}} \right)
the slope of the tangent line at the point (1,e)\left( {1,e} \right) is:
where x0=1{x_0} = 1, y0=e{y_0} = e and m=em = e, then
⇒y−e=e(x−1)\Rightarrow y - e = e\left( {x - 1} \right)
⇒y−e=ex−e\Rightarrow y - e = ex - e
Add both side by ee, then
⇒y−e+e=ex−e+e\Rightarrow y - e + e = ex - e + e
On simplification, we get
⇒y=ex\Rightarrow y = ex
Hence, the equation of the tangent line to the graph y=x2ex−2xex+2exy = {x^2}{e^x} - 2x{e^x} + 2{e^x} through the point (1,e)\left( {1,e} \right) is y=exy = ex.

Note : The concept of the equation of tangent comes under the concept of application of derivatives. Here the differentiation is applied on the equation of line or curve and then substitute the value of x. We should know about the general equation of a line. Hence these types of problems are solved by the above procedure.