What is the tangent of f(x)=sin(x),\at x=1 ?

The tangent of f(x)=sin(x),\at x=1 is y=cos(1)x+sin(1)-cos(1)

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tangent of f(x)=sin(x),\at x=1

tangent of

f (x) = sin (x), at x = 1

y = cos (1) x + sin (1) - cos (1)

Solution steps

Find the tangent point:

(1, sin (1))

Find the slope of

f (x) = sin (x) : \frac{df ( x )}{d 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 = cos (1)

Find the line with slope m=

cos (1)

and passing through

(1, sin (1)) : y = cos (1) x + sin (1) - cos (1)

y = cos (1) x + sin (1) - cos (1)

\int \frac{x ^{3} + 1}{x ^{2}} d x

\frac{d}{d x} (\frac{3}{x} + 5)

\int (\frac{1}{x ln ( x )}) d x

\frac{\partial}{\partial x} (x y^{2} - 5 y + 6)

f (x) = 6 x sin (x), at x = \frac{π}{2}

What is the tangent of f(x)=sin(x),\at x=1 ?
The tangent of f(x)=sin(x),\at x=1 is y=cos(1)x+sin(1)-cos(1)