Find the derivative. Simplify where possible.y = x coth(1 + x^2)

1 Answer |Add Yours

Posted on

You need to use definition of hyperbolic function coth x such that:

`coth x = (e^x + e^(-x))/(e^x- e^(-x))`

Hence,`y = x(e^(1+x^2) + e^(-1-x^2))/(e^(1+x^2) - e^(-1-x^2)).` You need to use product rule and chain rule to differentiate the aboveÂ equation with respect to x such that:

`(dy)/(dx) = x'*(e^(1+x^2) + e^(-1-x^2))/(e^(1+x^2) - e^(-1-x^2)) + x*((e^(1+x^2) + e^(-1-x^2))/(e^(1+x^2) - e^(-1-x^2)))'`

`(dy)/(dx) = (e^(1+x^2) + e^(-1-x^2))/(e^(1+x^2) - e^(-1-x^2)) + x*(((e^(1+x^2) + e^(-1-x^2))'*(e^(1+x^2) - e^(-1-x^2)) - *((e^(1+x^2) + e^(-1-x^2))*(e^(1+x^2) - e^(-1-x^2))'))/((e^(1+x^2) - e^(-1-x^2))^2)`

`(dy)/(dx) = (e^(1+x^2) + e^(-1-x^2))/(e^(1+x^2) - e^(-1-x^2)) + x*(((2xe^(1+x^2)- 2xe^(-1-x^2))*(e^(1+x^2) - e^(-1-x^2)) - ((e^(1+x^2) + e^(-1-x^2))*(2xe^(1+x^2)+ 2xe^(-1-x^2))))/((e^(1+x^2) - e^(-1-x^2))^2)`

`(dy)/(dx) = (e^(1+x^2) + e^(-1-x^2))/(e^(1+x^2) - e^(-1-x^2)) + 2x^2*(((e^(1+x^2) - e^(-1-x^2))*(e^(1+x^2) - e^(-1-x^2)) - ((e^(1+x^2) + e^(-1-x^2))*(e^(1+x^2) + e^(-1-x^2))))/((e^(1+x^2) - e^(-1-x^2))^2)`

`(dy)/(dx) = (e^(1+x^2) + e^(-1-x^2))/(e^(1+x^2) - e^(-1-x^2)) + 2x^2*(((e^(1+x^2) - e^(-1-x^2))*(e^(1+x^2) - e^(-1-x^2)) - ((e^(1+x^2) + e^(-1-x^2))*(e^(1+x^2) + e^(-1-x^2))))/((e^(1+x^2) - e^(-1-x^2))^2)`

`(dy)/(dx) = (e^(1+x^2) + e^(-1-x^2))/(e^(1+x^2) - e^(-1-x^2)) + 2x^2*(((e^(1+x^2) - e^(-1-x^2))^2 - ((e^(1+x^2) + e^(-1-x^2))^2)))/((e^(1+x^2) - e^(-1-x^2))^2)`

`(dy)/(dx) = (e^(1+x^2) + e^(-1-x^2))/(e^(1+x^2) - e^(-1-x^2)) + 2x^2*((e^(1+x^2) - e^(-1-x^2) - e^(1+x^2)- e^(-1-x^2))(e^(1+x^2) - e^(-1-x^2)+ e^(1+x^2)+ e^(-1-x^2)))/((e^(1+x^2) - e^(-1-x^2))^2)`

`(dy)/(dx) = (e^(1+x^2) + e^(-1-x^2))/(e^(1+x^2) - e^(-1-x^2))- 8x^2*((e^(-1-x^2))(e^(1+x^2)))/((e^(1+x^2) - e^(-1-x^2))^2)`

Hence, evaluating derivative of function yields `(dy)/(dx) = (e^(1+x^2) + e^(-1-x^2))/(e^(1+x^2) - e^(-1-x^2)) - 8x^2*((e^(-1-x^2))(e^(1+x^2)))/((e^(1+x^2) - e^(-1-x^2))^2)`

We’ve answered 323,597 questions. We can answer yours, too.