What is the tangent of 7+5x+xe^x ?

The tangent of 7+5x+xe^x is y=(e^{a_{0}}a_{0}+e^{a_{0}}+5)x-a_{0}^2e^{a_{0}}+7

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tangent of 7+5x+xe^x

tangent of

7 + 5 x + x e^{x}

y = (e^{a_{0}} a_{0} + e^{a_{0}} + 5) x - a_{0}^{2} e^{a_{0}} + 7

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 7 + 5 a_{0} + a_{0} e^{a_{0}})

Find the slope of

y = 7 + 5 x + x e^{x} : \frac{d y}{d x} = e^{x} x + e^{x} + 5

Compute the slope of

y = 7 + 5 x + x e^{x}

x = a_{0} : m = e^{a_{0}} a_{0} + e^{a_{0}} + 5

Find the line with slope m=

e^{a_{0}} a_{0} + e^{a_{0}} + 5

and passing through

(a_{0}, 7 + 5 a_{0} + a_{0} e^{a_{0}}) : y = (e^{a_{0}} a_{0} + e^{a_{0}} + 5) x - a_{0}^{2} e^{a_{0}} + 7

y = (e^{a_{0}} a_{0} + e^{a_{0}} + 5) x - a_{0}^{2} e^{a_{0}} + 7

x \to 8 + lim (\frac{x - 9}{x - 8})

\frac{\partial}{\partial x} (a e^{a x + y})

\frac{d}{d y} (\frac{cy ( x )}{x ^{2} + y ^{2}})

\int \frac{1}{tan ^{2} ( x ) + 1} d x

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (1) = 0, y^{^{'}} (1) = 1

What is the tangent of 7+5x+xe^x ?
The tangent of 7+5x+xe^x is y=(e^{a_{0}}a_{0}+e^{a_{0}}+5)x-a_{0}^2e^{a_{0}}+7