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need help with derivatives please :)?
1- use product and quotient rule to fid f'(x)
f(x)= [(2x+3)(3x-4)]/(x^2+3)
simplify your answer.
2- use chain rule to find f'(x)
f(x)=(x^2+1/x)^3/2
2 Answers
- Anonymous7 years ago
1.
let 2x+3 = a
3x-4 = b
x^2+3 =c
ab/c
on diff wrt x
(c ddx ab - ab ddx c)/ c^2
and ddx of ab is a ddx b + b ddx a
now put the values of a b c and solve.
2.
x^2 + 1/x = a
diff wrt x
3/2 a^(3/2-1) * ddx a
and ddx a is
ddx x^2 + ddx 1/x = 2x + log x
then solve after putting value
- Anonymous7 years ago
"Given f(x)= [(2x+3)(3x-4)]/(x^2+3)
=>f'(x)=[(d/dx(2x+3)(3x-4))*(x^2+3)-(d/dx(x^2+3))*(2x+3)(3x-4)]/(x^2+3)^2
=>f'(x)=[(12x+1)(x^2+3)-(2x)(2x+3)(3x-4)]/(x^2+3)^2
=>f'(x)=[3-(x-60)x]/(x^2+3)^2
Given f(x)=(x^2+1/x)^3/2
=>f'(x)=3/2(x^2+1/x)^(3/2-1)*2x*d/dx(1/x)
=>f'(x)=3/2√(x^2+1/x) (2x-1/x^2)"