Asked by Tima
Find the derivatives, dy/dx for the following functions.
a. x^2y^4 = lnx + 3y
b. y = ln(2x^2 + 1)
c. y = ln(ãx)
d. y = lnx/x
e. y = x^2 lnx
f. y ß log_2(3x)
a. x^2y^4 = lnx + 3y
b. y = ln(2x^2 + 1)
c. y = ln(ãx)
d. y = lnx/x
e. y = x^2 lnx
f. y ß log_2(3x)
Answers
Answered by
Steve
a. use implicit differentiation to get
2xy^4 + 4x^2 y^3 y' = 1/x + 3y'
y' = (1/x - 2xy^4)/(4x^2y^3 - 3)
= (1-2x^2y^4)/(4x^3y^3 - 3x)
b. use the chain rule
4x/(2x^2 + 1)
c. same
(1/√x)(1/2√x) = 1/(2x)
or, recognizing that ln(√x) = 1/2 lnx, (1/2)(1/x) = 1/(2x)
d. use the quotient rule
[(1/x)(x) - (lnx)(1)]/x^2
= (1-lnx)/x^2
e. product rule
2x lnx + x^2(1/x) = 2x lnx + x
f. y=log_2(3x)
= ln(3x)/ln(2)
y' = 3/(3x) / ln(2) = 1/(x ln2)
2xy^4 + 4x^2 y^3 y' = 1/x + 3y'
y' = (1/x - 2xy^4)/(4x^2y^3 - 3)
= (1-2x^2y^4)/(4x^3y^3 - 3x)
b. use the chain rule
4x/(2x^2 + 1)
c. same
(1/√x)(1/2√x) = 1/(2x)
or, recognizing that ln(√x) = 1/2 lnx, (1/2)(1/x) = 1/(2x)
d. use the quotient rule
[(1/x)(x) - (lnx)(1)]/x^2
= (1-lnx)/x^2
e. product rule
2x lnx + x^2(1/x) = 2x lnx + x
f. y=log_2(3x)
= ln(3x)/ln(2)
y' = 3/(3x) / ln(2) = 1/(x ln2)
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