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}}

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}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : 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}})

a re a y = sin (x), y = cos (x), x = \frac{5 π}{4}, x = 2 π

t an g e n t y = \frac{2 x}{x + 2}, (2, 1)

\frac{d}{d x} ((4^{x})^{9})

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}}