Question

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

Answer 1

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.

Explanation:

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))`