How do you differentiate #y=log_5(sqrt(3x+1))#?

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Answer 1 · 2017-04-11T09:25:43

Convert to the natural logarithm using, #log_5(A)=1/ln(5)ln(A)#, remove the square root by multiplying by #1/2#, then use the chain rule.

Given: #y=log_5(sqrt(3x+1))#

Convert to the natural logarithm:

#y=1/ln(5)ln(sqrt(3x+1))#

Remove the square root by multiplying by #1/2#:

#y=1/(2ln(5))ln|3x+1|#

Use the chain rule:

#dy/dx = (d(f(g(x))))/dx = (df)/(dg)(dg)/(dx)#

let #g = 3x+1#

let #f= 1/ln(5)ln(g)#

#(dg)/dx = 3#
#(df)/(dg)= 1/(2ln(5))1/g#

Substitute the above 2 equations into the chain rule:

#(d(f(g(x))))/dx = 1/(2ln(5))1/g3#

Reverse the substitution for g:

#(d(f(g(h(x)))))/dx = 1/(2ln(5))1/(3x+1)3#

#(d(f(g(h(x)))))/dx = 3/(2ln(5)(3x+1))#

#(d(f(g(x))))/dx = 3/((6x+2)ln(5))#

Use the property of logarithms that multiplication is the same as exponentiation within the argument:

#dy/dx = 3/ln(5^(6x+2))#