What is the tangent of x^4+7e^x ?

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

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tangent of x^4+7e^x

tangent of

x^{4} + 7 e^{x}

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

Find the slope of

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

Compute the slope of

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

x = a_{0} : m = 4 a_{0}^{3} + 7 e^{a_{0}}

Find the line with slope m=

4 a_{0}^{3} + 7 e^{a_{0}}

and passing through

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

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

n = 0 \sum \infty 9 (\frac{5}{12})^{n}

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

f (x) = (ln (x))^{2}, at (x = 11)

n = 1 \sum \infty 2 n

\frac{d y}{d x} - (2 + y) x^{2} = 0

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