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Chapter 16: Problem 13

Find the equation of the tangent to \(y=x \mathrm{e}^{x}\) where \(x=1\).

Short Answer

Expert verified
Answer: The equation of the tangent line is \(y = 2e(x - 1) + e\).

Step by step solution

01

Find the point on the curve where \(x=1\)

To find the point on the curve \(y = xe^x\) when \(x=1\), substitute \(x=1\) into the equation and solve for \(y\): $$ y = (1)e^{(1)} = e $$ So, the point on the curve is \((1, e)\).
02

Find the derivative of the function

To find the slope of the tangent at the point \((1, e)\), we need to find the derivative of \(y = xe^x\). We can do this by applying the product rule, which is given by: $$ \frac{d(uv)}{dx} = u\frac{dv}{dx} + v\frac{du}{dx} $$ In our case, let \(u=x\) and \(v=e^x\). Then, \(\frac{du}{dx} = 1\) and \(\frac{dv}{dx} = e^x\). Now apply the product rule: $$ \frac{dy}{dx} = x\frac{d(e^x)}{dx} + e^x\frac{d(x)}{dx} = xe^x + e^x $$
03

Find the slope of the tangent at the point \((1,e)\)

Now, we need to find the slope of the tangent line at the point \((1, e)\). We do this by evaluating the derivative at \(x=1\): $$ \frac{dy}{dx}(1) = (1)e^1 + e^1 = e + e = 2e $$ Thus, the slope of the tangent line at the point \((1, e)\) is \(2e\).
04

Write the equation of the tangent line

We now have the point and the slope of the tangent line, namely \((1, e)\) and \(2e\), respectively. To find the equation of the tangent, we can use the point-slope form of a linear equation, given by: $$ y - y_1 = m(x - x_1) $$ Substituting the point \((1, e)\) and the slope \(2e\) into this equation, we have: $$ y - e = 2e(x - 1) $$ To make it explicit, we can rewrite the equation as: $$ y = 2e(x - 1) + e $$ This is the equation of the tangent line to the curve \(y = xe^x\) at the point \((1, e)\).

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Most popular questions from this chapter

State the range of values of \(x\) for which each of the following functions is (i) concave up (ii) concave down: (a) \(y=\frac{x^{3}}{6}-\frac{5 x^{2}}{2}+3 x-9\) (b) \(y=3+x+\frac{x^{2}}{2}-\frac{x^{4}}{12}\) (c) \(y=\mathrm{e}^{x}-100 x-100\) (d) \(y=(x-1)^{4}\) (e) \(y=\frac{x^{2}}{2} \ln x-\frac{x^{2}}{4}\)

Find the derivative of the following functions: (a) \(y=\mathrm{e}^{2 x} x^{3} \sin 3 x \cos 2 x\) (b) \(y=(x+\sin x)^{7}\) (c) \(H=\ln \left(t^{2}+3 t-9\right)\) (d) \(V(r)=\frac{1}{\ln r}\) (e) \(M(b)=\ln b+\ln (b+1)\)

Find \(y^{\prime}\) where \(y\) is given by (a) \(\frac{x^{2}+1}{2 x+3}\) (b) \(\frac{\cos 2 t}{\sin 3 t}\) (c) \(\frac{r^{3}}{3 e^{2 r}}\) (d) \(\frac{v+\sin v}{1+\mathrm{e}^{v}}\) (e) \(\frac{\sqrt{x}}{x+1}\)

Calculate \(y^{\prime}\) where \(y\) is given by (a) \(\left(t^{2}+1\right) \sin 4 t\) (b) \((3 t+7) \mathrm{e}^{-2 t}\) (c) \(\left(e^{x}+\mathrm{e}^{-2 \mathrm{n}}\right)\left(3 x^{2}-2 x\right)\) (d) \(\sqrt{x} e^{x}\) (e) \(\frac{t^{2}+1}{\mathrm{e}^{t}}\)

Find \(\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}\) given \(y=\mathrm{e}^{\left(x^{2}\right)}\).
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