What is the tangent of y=3+4x+7xe^x ?

The tangent of y=3+4x+7xe^x is y=(7(e^{a_{0}}a_{0}+e^{a_{0}})+4)x-7a_{0}^2e^{a_{0}}+3

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tangent of y=3+4x+7xe^x

tangent of

y = 3 + 4 x + 7 x e^{x}

y = (7 (e^{a_{0}} a_{0} + e^{a_{0}}) + 4) x - 7 a_{0}^{2} e^{a_{0}} + 3

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 3 + 4 a_{0} + 7 a_{0} e^{a_{0}})

Find the slope of

y = 3 + 4 x + 7 x e^{x} : \frac{d y}{d x} = 7 (e^{x} x + e^{x}) + 4

Compute the slope of

y = 3 + 4 x + 7 x e^{x}

x = a_{0} : m = 7 (e^{a_{0}} a_{0} + e^{a_{0}}) + 4

Find the line with slope m=

7 (e^{a_{0}} a_{0} + e^{a_{0}}) + 4

and passing through

(a_{0}, 3 + 4 a_{0} + 7 a_{0} e^{a_{0}}) : y = (7 (e^{a_{0}} a_{0} + e^{a_{0}}) + 4) x - 7 a_{0}^{2} e^{a_{0}} + 3

y = (7 (e^{a_{0}} a_{0} + e^{a_{0}}) + 4) x - 7 a_{0}^{2} e^{a_{0}} + 3

x^{2} y^{^{''}} + 3 x y^{^{'}} - 35 y = 0

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\int \frac{8 x}{1 + x ^{2}} d x

\frac{d y}{d x} + y^{5} x + 6 y = 0

\int \frac{1}{x ^{2} 4 - x ^{2}} d x

What is the tangent of y=3+4x+7xe^x ?
The tangent of y=3+4x+7xe^x is y=(7(e^{a_{0}}a_{0}+e^{a_{0}})+4)x-7a_{0}^2e^{a_{0}}+3