对于一个连续的函数,切点处的导数等于切线斜率。我们只需要知道切点坐标和切点斜率,就能求切线方程。
过( 1 , e ) (1,e)(1 ,e )作y = e x y=e^x y =e x的切线,求切线方程。
解:
y ′ = e x ∣ x = 1 = e \qquad y’=e^x|_{x=1}=e y ′=e x ∣x =1 =e
\qquad所以k = e k=e k =e
y − e = e ( x − 1 ) \qquad y-e=e(x-1)y −e =e (x −1 ),即y = e x y=ex y =e x
y = 3 ( x 2 + x ) e x y=3(x^2+x)e^x y =3 (x 2 +x )e x在点( 0 , 0 ) (0,0)(0 ,0 )处的切线方程为‾ \underline{\qquad}
解:
y ′ = 3 ( x 2 + 3 x + 1 ) e x ∣ x = 0 = 3 \qquad y’=3(x^2+3x+1)e^x|_{x=0}=3 y ′=3 (x 2 +3 x +1 )e x ∣x =0 =3
\qquad所以k = 3 k=3 k =3
( y − 0 ) = 3 ( x − 0 ) \qquad (y-0)=3(x-0)(y −0 )=3 (x −0 ),即y = 3 x y=3x y =3 x
过( 0 , 1 ) (0,1)(0 ,1 )作y = ln x y=\ln x y =ln x的切线,求切线方程。
解:
\qquad设切点为( x 0 , ln x 0 ) (x_0,\ln x_0)(x 0 ,ln x 0 ),则y ′ = 1 x ∣ x = x 0 = 1 x 0 y’=\dfrac 1x|_{x=x_0}=\dfrac{1}{x_0}y ′=x 1 ∣x =x 0 =x 0 1
\qquad所以k = 1 x 0 k=\dfrac{1}{x_0}k =x 0 1 ,即y − ln x 0 = 1 x 0 ( x − x 0 ) y-\ln x_0=\dfrac{1}{x_0}(x-x_0)y −ln x 0 =x 0 1 (x −x 0 )
\qquad将( 0 , 1 ) (0,1)(0 ,1 )代入得1 − ln x 0 = 1 x 0 ( 0 − x 0 ) 1-\ln x_0=\dfrac{1}{x_0}(0-x_0)1 −ln x 0 =x 0 1 (0 −x 0 ),解得x 0 = e 2 x_0=e^2 x 0 =e 2
\qquad将x 0 x_0 x 0 代入y = ln x y=\ln x y =ln x,得y 0 = 2 y_0=2 y 0 =2,切点为( e 2 , 2 ) (e^2,2)(e 2 ,2 )
\qquad切线为y − 2 = 1 e 2 ( x − e 2 ) y-2=\dfrac{1}{e^2}(x-e^2)y −2 =e 2 1 (x −e 2 ),即y = 1 e 2 x + 1 y=\dfrac{1}{e^2}x+1 y =e 2 1 x +1
本篇文章简单讲了用导数求切线方程中给切点和给切线上一点的解题方法,只做了解,并没有深入探讨,目的是让大家更多地了解导数的用途。
Original: https://blog.csdn.net/tanjunming2020/article/details/127825222
Author: tanjunming2020
Title: 导数求切线
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