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Find the derivative of the function. `y=sqrt(7+ lnz)`

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sllysmith | Student, Undergraduate | eNoter

Posted March 15, 2012 at 5:41 PM via web

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Find the derivative of the function. `y=sqrt(7+ lnz)`

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txmedteach | High School Teacher | (Level 3) Associate Educator

Posted March 15, 2012 at 8:24 PM (Answer #1)

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To calculte the derivative of a function like this, you must use the chain rule, seen here:

`dy/dz = (dy)/(du) * (du)/dz`

What this rule implies is that we will define an intermediate function, `u`, such that we can take the derivative of `y` with respect to this function and so that we can take the derivative of this function with respect to `z`. The derivative of `y` with respect to `z` will be the product of these results. Also, notice that `du` cancels out in the above product.

In our case, a convenient `u` would be as follows:

`u = 7+lnz`

We can easily find `(du)/dz`:

`(du)/(dz) = 1/z`

Now, we must first find `y` in terms of `u`:

`y = sqrtu`

Now, we must differentiate, noting that `sqrtu = u^(1/2)`:

`(dy)/(du) = 1/2*u^-(1/2) = 1/(2sqrtu)`

Now, we can solve for the overall derivative `dy/dz`:

`(dy)/(dz) = (dy)/(du)*(du)/(dz) = 1/(2sqrtu) * 1/z`

To find the result for this final expression, we must substitute `7+lnz` for `u`:

`(dy)/(dz) = 1/(2zsqrt(7+lnz))`

There is your final answer! I hope this helps!

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