回答:
说明:
我们需要使用两个规则:产品规则和链规则。产品规则规定:
链规则规定:
因此,
找到的衍生物
将此结果替换为原始等式:
什么是(sqrt(5+)sqrt(3))/(sqrt(3+)sqrt(3+)sqrt(5)) - (sqrt(5-)sqrt(3))/(sqrt(3+)sqrt (3-)SQRT(5))?
2/7我们采取,A =(sqrt5 + sqrt3)/(sqrt3 + sqrt3 + sqrt5) - (sqrt5-sqrt3)/(sqrt3 + sqrt3-sqrt5)=(sqrt5 + sqrt3)/(2sqrt3 + sqrt5) - (sqrt5 -sqrt3)/(2sqrt3-sqrt5)=(sqrt5 + sqrt3)/(2sqrt3 + sqrt5) - (sqrt5-sqrt3)/(2sqrt3-sqrt5)=((sqrt5 + sqrt3)(2sqrt3-sqrt5) - (sqrt5-sqrt3) )(2sqrt3 + sqrt5))/((2sqrt3 + sqrt5)(2sqrt3-sqrt5)=((2sqrt15-5 + 2 * 3-sqrt15) - (2sqrt15 + 5-2 * 3-sqrt15))/((2sqrt3) ^ 2-(sqrt5)^ 2)=(取消(2sqrt15)-5 + 2 * 3cancel(-sqrt15) - 取消(2sqrt15)-5 + 2 * 3 +取消(sqrt15))/(12-5)=( -10 + 12)/ 7 = 2/7请注意,如果在分母中(sqrt3 + sqrt(3 + sqrt5))和(sqrt3 + sqrt(3-sqrt5))那么答案将会改变。
F(x)= sqrt(1 + log_3(x)?的导数是多少?
D / dx(sqrt(1 + log_3x))=((d / dx)(1 + log_3x))/ {2sqrt(1 + log_3x)} =((d / dx)(1 + logx / log3))/ { 2sqrt(1 + log_3x)} =(1 /(xln3))/ {2sqrt(1 + log_3x)} = 1 /(2xln3sqrt(1 + log_3))
你如何简化(1 / sqrt(a-1)+ sqrt(a + 1))/(1 / sqrt(a + 1)-1 / sqrt(a-1))div sqrt(a + 1)/( (a-1)sqrt(a + 1) - (a + 1)sqrt(a-1)),a> 1?
巨大的数学格式......>颜色(蓝色)(((1 / sqrt(a-1)+ sqrt(a + 1))/(1 / sqrt(a + 1)-1 / sqrt(a-1)) )/(sqrt(a + 1)/((a-1)sqrt(a + 1) - (a + 1)sqrt(a-1)))=颜色(红色)(((1 / sqrt(a-) 1)+ sqrt(a + 1))/((sqrt(a-1)-sqrt(a + 1))/(sqrt(a + 1)cdot sqrt(a-1))))/(sqrt(a) +1)/(sqrt(a-1)cdot sqrt(a-1)cdot sqrt(a + 1)-sqrt(a + 1)cdot sqrt(a + 1)sqrt(a-1)))= color(蓝色)(((1 / sqrt(a-1)+ sqrt(a + 1))/((sqrt(a-1)-sqrt(a + 1))/(sqrt(a + 1)cdot sqrt(a -1))))/(sqrt(a + 1)/(sqrt(a + 1)cdot sqrt(a-1)(sqrt(a-1)-sqrt(a + 1)))=颜色(红色) ((1 / sqrt(a-1)+ sqrt(a + 1))/((sqrt(a-1)-sqrt(a + 1))/(sqrt(a + 1)cdot sqrt(a-1) ))xx(sqrt(a + 1)cdot sqrt(a-1)(sqrt(a-1)-sqrt(a + 1))