What is the tangent of x^4+9e^x ?

The tangent of x^4+9e^x is y=(4a_{0}^3+9e^{a_{0}})x-3a_{0}^4-9a_{0}e^{a_{0}}+9e^{a_{0}}

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tangent of x^4+9e^x

tangent of

x^{4} + 9 e^{x}

y = (4 a_{0}^{3} + 9 e^{a_{0}}) x - 3 a_{0}^{4} - 9 a_{0} e^{a_{0}} + 9 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} + 9 e^{a_{0}})

Find the slope of

y = x^{4} + 9 e^{x} : \frac{d y}{d x} = 4 x^{3} + 9 e^{x}

Compute the slope of

y = x^{4} + 9 e^{x}

x = a_{0} : m = 4 a_{0}^{3} + 9 e^{a_{0}}

Find the line with slope m=

4 a_{0}^{3} + 9 e^{a_{0}}

and passing through

(a_{0}, a_{0}^{4} + 9 e^{a_{0}}) : y = (4 a_{0}^{3} + 9 e^{a_{0}}) x - 3 a_{0}^{4} - 9 a_{0} e^{a_{0}} + 9 e^{a_{0}}

y = (4 a_{0}^{3} + 9 e^{a_{0}}) x - 3 a_{0}^{4} - 9 a_{0} e^{a_{0}} + 9 e^{a_{0}}

\frac{d y}{d x}, y = e^{x}

\frac{d y}{d t} - \frac{2 t}{1 + t ^{2}} y = 3

tan^{2} (y)

\frac{\partial}{\partial x} (2 x^{3} + 3 y^{2} - 5 x y)

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

What is the tangent of x^4+9e^x ?
The tangent of x^4+9e^x is y=(4a_{0}^3+9e^{a_{0}})x-3a_{0}^4-9a_{0}e^{a_{0}}+9e^{a_{0}}