What is the tangent of f(x)=e^xcos(x),(0,1) ?

The tangent of f(x)=e^xcos(x),(0,1) is y=x+1

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tangent of f(x)=e^xcos(x),(0,1)

tangent of

f (x) = e^{x} cos (x), (0, 1)

y = x + 1

Solution steps

Find the slope of

f (x) = e^{x} cos (x) : \frac{df ( x )}{d x} = e^{x} cos (x) - e^{x} sin (x)

Compute the slope of

f (x) = e^{x} cos (x)

(0, 1) : m = 1

Find the line with slope m=

1

and passing through

(0, 1) : y = x + 1

y = x + 1

\frac{\partial}{\partial x} (ln (7 x^{2} + y^{2} + 6))

8 x^{3}

x y^{^{'}} + 2 y = - 2 e^{x}

x \to - 3 lim (\frac{7}{x + 3})

y^{^{'}} = t^{2} - y

What is the tangent of f(x)=e^xcos(x),(0,1) ?
The tangent of f(x)=e^xcos(x),(0,1) is y=x+1