What is the tangent of 7+2e^x-4x^4x-y=9 ?

The tangent of 7+2e^x-4x^4x-y=9 is y=(-20a_{0}^4+2e^{a_{0}})x+16a_{0}^5-2a_{0}e^{a_{0}}+2e^{a_{0}}-2

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tangent of 7+2e^x-4x^4x-y=9

tangent of

7 + 2 e^{x} - 4 x^{4} x - y = 9

y = (- 20 a_{0}^{4} + 2 e^{a_{0}}) x + 16 a_{0}^{5} - 2 a_{0} e^{a_{0}} + 2 e^{a_{0}} - 2

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, - 4 a_{0}^{5} - 2 + 2 e^{a_{0}})

Find the slope of

7 + 2 e^{x} - 4 x^{4} x - y = 9 : \frac{d y}{d x} = - 20 x^{4} + 2 e^{x}

Compute the slope of

7 + 2 e^{x} - 4 x^{4} x - y = 9

x = a_{0} : m = - 20 a_{0}^{4} + 2 e^{a_{0}}

Find the line with slope m=

- 20 a_{0}^{4} + 2 e^{a_{0}}

and passing through

(a_{0}, - 4 a_{0}^{5} - 2 + 2 e^{a_{0}}) : y = (- 20 a_{0}^{4} + 2 e^{a_{0}}) x + 16 a_{0}^{5} - 2 a_{0} e^{a_{0}} + 2 e^{a_{0}} - 2

y = (- 20 a_{0}^{4} + 2 e^{a_{0}}) x + 16 a_{0}^{5} - 2 a_{0} e^{a_{0}} + 2 e^{a_{0}} - 2

\frac{\partial}{\partial x} (x^{2 x})

t \to 1 lim (\frac{t ^{3} - 1}{t - 1})

x \to + 0 - lim (\frac{6}{5 x ^{\frac{1}{3}}})

\frac{d}{d x} (sin (x) cos^{2} (x))

\int \frac{1}{4 - 4 sin ( x )} d x

What is the tangent of 7+2e^x-4x^4x-y=9 ?
The tangent of 7+2e^x-4x^4x-y=9 is y=(-20a_{0}^4+2e^{a_{0}})x+16a_{0}^5-2a_{0}e^{a_{0}}+2e^{a_{0}}-2