# Find y' if `y = (ln(2x+1)^5)/(sqrt(x^2+1))`

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### 1 Answer

It is given that `y=(ln(2x+1)^5)/(sqrt(x^2+1))`

The derivative of y can be found by using the chain rule and the quotient rule.

`y' = ((ln(2x+1)^5)'*sqrt(x^2+1) - ln(2x+1)^5*(sqrt(x^2+1))')/(x^2 + 1)`

=> `(10/(2x + 1)*sqrt(x^2+1) - 5*ln(2x + 1)*(1/2)*2x/sqrt(x^2+1))/(x^2 + 1)`

=> `(10*sqrt(x^2 + 1)/(2x + 1) - (5x*ln(2x+1))/sqrt(x^2 + 1))/(x^2 + 1)`

**The required derivative is ** `y' = (10*sqrt(x^2 + 1)/(2x + 1) - (5x*ln(2x+1))/sqrt(x^2 + 1))/(x^2 + 1)`