What is the tangent of f(x)=x^2+sin(x) ?

The tangent of f(x)=x^2+sin(x) is y=(2a_{0}+cos(a_{0}))x-a_{0}^2-a_{0}cos(a_{0})+sin(a_{0})

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tangent of f(x)=x^2+sin(x)

tangent of

f (x) = x^{2} + sin (x)

y = (2 a_{0} + cos (a_{0})) x - a_{0}^{2} - a_{0} cos (a_{0}) + sin (a_{0})

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{2} + sin (a_{0}))

Find the slope of

f (x) = x^{2} + sin (x) : \frac{df ( x )}{d x} = 2 x + cos (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 = 2 a_{0} + cos (a_{0})

Find the line with slope m=

2 a_{0} + cos (a_{0})

and passing through

(a_{0}, a_{0}^{2} + sin (a_{0})) : y = (2 a_{0} + cos (a_{0})) x - a_{0}^{2} - a_{0} cos (a_{0}) + sin (a_{0})

y = (2 a_{0} + cos (a_{0})) x - a_{0}^{2} - a_{0} cos (a_{0}) + sin (a_{0})

\int x^{2} - 3 x d x

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

\frac{d}{d x} (lo g_{e} (x + 1))

\int \frac{1}{cos ( θ ) - 1} d θ

y^{^{'''}} - 9 y^{^{''}} + 15 y^{^{'}} + 25 y = 0

What is the tangent of f(x)=x^2+sin(x) ?
The tangent of f(x)=x^2+sin(x) is y=(2a_{0}+cos(a_{0}))x-a_{0}^2-a_{0}cos(a_{0})+sin(a_{0})