What is the tangent of y=sin(t)+1/6 e^t,(pi, 1/6 e^pi) ?

The tangent of y=sin(t)+1/6 e^t,(pi, 1/6 e^pi) is y=(-1+(e^pi)/6)t+(e^pi+pi(6-e^pi))/6

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tangent of y=sin(t)+1/6 e^t,(pi, 1/6 e^pi)

tangent of

y = sin (t) + \frac{1}{6} e^{t}, (π, \frac{1}{6} e^{π})

y = (- 1 + \frac{e ^{π}}{6}) t + \frac{e ^{π} + π ( 6 - e ^{π} )}{6}

Solution steps

Find the slope of

y = sin (t) + \frac{1}{6} e^{t} : \frac{d y}{d t} = cos (t) + \frac{1}{6} e^{t}

Compute the slope of

y = sin (t) + \frac{1}{6} e^{t}

(π, \frac{1}{6} e^{π}) : m = - 1 + \frac{e ^{π}}{6}

Find the line with slope m=

- 1 + \frac{e ^{π}}{6}

and passing through

(π, \frac{1}{6} e^{π}) : y = (- 1 + \frac{e ^{π}}{6}) t + \frac{e ^{π} + π ( 6 - e ^{π} )}{6}

y = (- 1 + \frac{e ^{π}}{6}) t + \frac{e ^{π} + π ( 6 - e ^{π} )}{6}

\frac{d}{d x} y = x^{4} \cdot x^{5}

\int \frac{( x ^{2} + 1 )}{( x - 8 ) ( x - 7 ) ^{2}} d x

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

\frac{\partial}{\partial x} (\frac{1 + x}{x + y})

\int t^{2} + t d t

What is the tangent of y=sin(t)+1/6 e^t,(pi, 1/6 e^pi) ?
The tangent of y=sin(t)+1/6 e^t,(pi, 1/6 e^pi) is y=(-1+(e^pi)/6)t+(e^pi+pi(6-e^pi))/6