What is the tangent of x^4+2e^x ?

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

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tangent of x^4+2e^x

tangent of

x^{4} + 2 e^{x}

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{4} + 2 e^{a_{0}})

Find the slope of

y = x^{4} + 2 e^{x} : \frac{d y}{d x} = 4 x^{3} + 2 e^{x}

Compute the slope of

y = x^{4} + 2 e^{x}

x = a_{0} : m = 4 a_{0}^{3} + 2 e^{a_{0}}

Find the line with slope m=

4 a_{0}^{3} + 2 e^{a_{0}}

and passing through

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

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

\int sin (24 x) d x

\int_{0}^{2} 1 + x d x

\frac{x}{1 - \frac{1}{x ^{2}}}

\int 3 (3 x + 9)^{5} d x

(t^{2} + 2 t) \frac{d x}{d t} = 2 x + 7, x (1) = 1

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