Question

高数问题，帮忙解答？

Answer 1

x=arctant
y=ln(1+t²)
dy/dx=(dy/dt)/(dx/dt)
=[2t/(1+t²)]/[1/(1+t²)]
=2t

x=1+t²
y=t³
t=2时，x=5，y=8
dy/dx=(dy/dt)/(dx/dt)=3t²/(2t)=3t/2
t=2处切线斜率=3/2·2=3
切线为y–8=3(x–5)
即y=3x–7

3.x=t²+2t
y=ln(1+t)
x=3时，t=1，y=ln2
dy/dx=(dy/dt)/(dx/dt)=[1/(1+t)]/(2t+2)
=1/[2(1+t)²]
x=3处切线斜率=1/8
法线斜率为–8
法线方程y–ln2=–8(x–3)
y=–8x+ln2+24
y=0时，x=3+(ln2)/8
与x轴交点坐标为(3+ln2 /8，0)

4.y=ln∨[(1–x)/(1+x²)]
y'=∨[(1+x²)/(1–x)] ·1/2·[(1–x)/(1+x²)]^(–1/2)
·[(1–x)/(1+x²)]'
=(1+x²)/[2(1–x)]·[–(1+x²)–(1–x)·2x]/(1+x²)²
=(x²–2x–1)/[2(1–x)(1+x²)]
=1/[2(1–x)]–(x+1)/[(1–x)(1+x²)]
y''=1/[2(1–x)²]–[(1–x)(1+x²)+(x+1)(1–2x+3x²)]/[(1–x)(1+x²)]²
x=0时，y''=1/2–(1+1)/1=–3/2

5.y=(sinx)^4–(cosx)^4
y'=4sin³xcosx+4cos³xsinx
=4sinxcosx(sin²x+cos²x)
=2sin2x
y''=4cos2x