What is the tangent of f(x)=sin(x)cos(x) ?

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

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tangent of f(x)=sin(x)cos(x)

tangent of

f (x) = sin (x) cos (x)

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, sin (a_{0}) cos (a_{0}))

Find the slope of

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

Find the line with slope m=

cos (2 a_{0})

and passing through

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

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

\int \frac{x}{( x ^{2} + 1 ) ( x ^{2} + 4 x + 4 )} d x

x \to 0 lim (8 x csc (5 x))

11 (t + 1) \frac{d y}{d t} - 8 y = 24 t

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

\frac{d}{d t} (t - \frac{1}{t})

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