What is the tangent of y=x^4+2e^x(0.2) ?

The tangent of y=x^4+2e^x(0.2) is y=(4a_{0}^3+0.4e^{a_{0}})x-3a_{0}^4-0.4a_{0}e^{a_{0}}+0.4e^{a_{0}}

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tangent of y=x^4+2e^x(0.2)

tangent of

y = x^{4} + 2 e^{x} (0.2)

y = (4 a_{0}^{3} + 0.4 e^{a_{0}}) x - 3 a_{0}^{4} - 0.4 a_{0} e^{a_{0}} + 0.4 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} + 0.4 e^{a_{0}})

Find the slope of

y = x^{4} + 2 e^{x} 0.2 : \frac{d y}{d x} = 4 x^{3} + 0.4 e^{x}

Compute the slope of

y = x^{4} + 2 e^{x} 0.2

x = a_{0} : m = 4 a_{0}^{3} + 0.4 e^{a_{0}}

Find the line with slope m=

4 a_{0}^{3} + 0.4 e^{a_{0}}

and passing through

(a_{0}, a_{0}^{4} + 0.4 e^{a_{0}}) : y = (4 a_{0}^{3} + 0.4 e^{a_{0}}) x - 3 a_{0}^{4} - 0.4 a_{0} e^{a_{0}} + 0.4 e^{a_{0}}

y = (4 a_{0}^{3} + 0.4 e^{a_{0}}) x - 3 a_{0}^{4} - 0.4 a_{0} e^{a_{0}} + 0.4 e^{a_{0}}

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

\frac{d}{d x} (cot^{2} (7 x))

\int \frac{k}{k - x} d x

\frac{d y}{d x} = \frac{x - 4}{y + 6}

\frac{d}{d x} (2 x + \frac{( x ^{2} - x )}{( x ^{3} + x + 1 )})

What is the tangent of y=x^4+2e^x(0.2) ?
The tangent of y=x^4+2e^x(0.2) is y=(4a_{0}^3+0.4e^{a_{0}})x-3a_{0}^4-0.4a_{0}e^{a_{0}}+0.4e^{a_{0}}