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tangent y=x^{sin(x)}

解答

切线

y = x^{s i n (x)}

解答

y = a_{0}^{s i n (a_{0})} (cos (a_{0}) ln (a_{0}) + \frac{sin ( a _{0} )}{a _{0}}) x + a_{0}^{s i n (a_{0})} - a_{0}^{s i n (a_{0}) + 1} (ln (a_{0}) cos (a_{0}) + \frac{sin ( a _{0} )}{a _{0}})

求解步骤

计算在一般点

x = a_{0}

的切线

找到切点:

(a_{0}, a_{0}^{s i n (a_{0})})

计算

y = x^{s i n (x)}

的斜率:

\frac{d y}{d x} = x^{s i n (x)} (cos (x) ln (x) + \frac{sin ( x )}{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 = a_{0}^{s i n (a_{0})} (cos (a_{0}) ln (a_{0}) + \frac{sin ( a _{0} )}{a _{0}})

找到斜率 m=

a_{0}^{s i n (a_{0})} (cos (a_{0}) ln (a_{0}) + \frac{sin ( a _{0} )}{a _{0}})

且穿过

(a_{0}, a_{0}^{s i n (a_{0})})

的直线:

y = a_{0}^{s i n (a_{0})} (cos (a_{0}) ln (a_{0}) + \frac{sin ( a _{0} )}{a _{0}}) x + a_{0}^{s i n (a_{0})} - a_{0}^{s i n (a_{0}) + 1} (ln (a_{0}) cos (a_{0}) + \frac{sin ( a _{0} )}{a _{0}})

y = a_{0}^{s i n (a_{0})} (cos (a_{0}) ln (a_{0}) + \frac{sin ( a _{0} )}{a _{0}}) x + a_{0}^{s i n (a_{0})} - a_{0}^{s i n (a_{0}) + 1} (ln (a_{0}) cos (a_{0}) + \frac{sin ( a _{0} )}{a _{0}})

作图

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