What is the derivative of #ln[(2x-3)/(7x+8)]^(1/2)#?

show below we will use the properties of ln

now the derivative

To find the derivative of ( \ln\left[\left(\frac{2x - 3}{7x + 8}\right)^{1/2}\right] ), you can use the chain rule and the properties of logarithmic differentiation. The derivative is:

[ \frac{d}{dx} \ln\left[\left(\frac{2x - 3}{7x + 8}\right)^{1/2}\right] = \frac{1}{\left[\left(\frac{2x - 3}{7x + 8}\right)^{1/2}\right]} \cdot \frac{d}{dx} \left(\frac{2x - 3}{7x + 8}\right)^{1/2} ]

Using the chain rule and the power rule for differentiation, the derivative of ( \left(\frac{2x - 3}{7x + 8}\right)^{1/2} ) is:

[ \frac{1}{2\sqrt{\frac{2x - 3}{7x + 8}}} \cdot \frac{d}{dx}\left(\frac{2x - 3}{7x + 8}\right) ]

Now, differentiate ( \frac{2x - 3}{7x + 8} ) with respect to ( x ):

[ \frac{d}{dx}\left(\frac{2x - 3}{7x + 8}\right) = \frac{(7x + 8)(2) - (2x - 3)(7)}{(7x + 8)^2} ]

Simplify this expression:

[ \frac{(14x + 16) - (14x - 21)}{(7x + 8)^2} = \frac{37}{(7x + 8)^2} ]

Putting it all together:

[ \frac{d}{dx} \ln\left[\left(\frac{2x - 3}{7x + 8}\right)^{1/2}\right] = \frac{1}{\left[\left(\frac{2x - 3}{7x + 8}\right)^{1/2}\right]} \cdot \frac{1}{2\sqrt{\frac{2x - 3}{7x + 8}}} \cdot \frac{37}{(7x + 8)^2} ]

Simplify the expression further if necessary.