设y=ln(x+根号下(x^2+a^2)),求dy.
人气:113 ℃ 时间:2019-09-06 05:10:59
解答
∵y=ln[x+√(x^2+a^2)],∴e^y=x+√(x^2+a^2),∴(e^y-x)^2=x^2+a^2,
∴2(e^y-x)(e^y-x)′=2x,∴[x+√(x^2+a^2)-x][(e^y)y′-1]=x,
∴(e^y)y′-1=x/√(x^2+a^2),
∴(e^y)y′=1+x/√(x^2+a^2)=[x+√(x^2+a^2)]/√(x^2+a^2),
∴y′=1/√(x^2+a^2),
∴dy=[1/√(x^2+a^2)]dx.
推荐
- ln[根号(x^2+y^2)] =arctany/x 求dy
- 由方程arctan y/x=ln(根号x^2+y^2)所确定的y是x的函数,求dy/dx
- 设函数y=ln(x+根号下x²+1)求微积分dy
- 1、lim(x趋近于0)xcotx-1/x平方.2、设y=ln根号1-x/arccosx,求y'(0).3、设y=ln(1+x+y),求dy/dx.
- 设y=根号下[x+(根号x)],求dy(1)=?
- 已知α,β∈(0,π/4),且3sinβ=sin(2α+β),4tanα/2=1-tanα/2,求α+β的值
- Do not want to say good-bye,but have to say
- All the neighbours admire this family,___the children and parents have built up a friendly relationship.这句话空格处答案是where,只
猜你喜欢