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# logarithmic derivative

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find derivative of y=log of base 7 (Square root(9x+9))

I have first use change of base formula 1/2 (ln(9x+9)/ ln7 but then what do I do?

thanks!

May 3, 2018

#1
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Find the derivative of the following via implicit differentiation:
d/dx(y) = d/dx((log(sqrt(9 + 9 x)))/log(7))

The derivative of y is y'(x):
y'(x) = d/dx((log(sqrt(9 + 9 x)))/log(7))

Factor out constants:
y'(x) = (d/dx(log(sqrt(9 + 9 x))))/(log(7))

Simplify log(sqrt(9 x + 9)) using the identity log(a^b) = b log(a):
y'(x) = (d/dx(1/2 log(9 + 9 x)))/log(7)

Factor out constants:
y'(x) = ((d/dx(log(9 + 9 x)))/2)/log(7)

Using the chain rule, d/dx(log(9 x + 9)) = ( dlog(u))/( du) ( du)/( dx), where u = 9 x + 9 and d/( du)(log(u)) = 1/u:
y'(x) = (d/dx(9 + 9 x))/(9 x + 9) 1/(2 log(7))

Simplify the expression:
y'(x) = (d/dx(9 + 9 x))/(2 (9 + 9 x) log(7))

Differentiate the sum term by term and factor out constants:
y'(x) = d/dx(9) + 9 d/dx(x) 1/(2 (9 + 9 x) log(7))
The derivative of 9 is zero:
y'(x) = (9 (d/dx(x)) + 0)/(2 (9 + 9 x) log(7))

Simplify the expression:
y'(x) = (9 (d/dx(x)))/(2 (9 + 9 x) log(7))

The derivative of x is 1:
y'(x) = 1 9/(2 (9 + 9 x) log(7))

Simplify the expression:
y'(x) = 9/(2 (9 + 9 x) log(7))

Factor the numerator and denominator of the right hand side:
y'(x) = 9/(18 (1 + x) log(7))

Cancel common terms in the numerator and denominator:

y'(x) = 1/(2 (1 + x) log(7))   [Courtesy of Mathematica 11 Home Edition]

May 3, 2018