Respuesta :
The partial derivative dy/dx at x = e is given as follows:
[tex]\frac{dy}{dx}(e) = -\frac{2}{e}[/tex]
How to obtain the partial derivative?
The function in this problem is defined as follows:
y = tan(u).
The function y is a function of the variable u, which is a function of the variable v, which is a function of variable x, then the function of variable x can be written as follows:
[tex]y = \tan{\left(\ln{x} - \frac{1}{\ln{x}}\right)}[/tex]
Then, applying the chain rule, the derivative is the derivative of the outer tangent function multiplied by the derivative of the inner ln function.
The inner function is of:
[tex]\ln{x} - \frac{1}{\ln{x}}[/tex]
Which has a derivative of:
[tex]\frac{1}{x} + \frac{1}{x\ln^2{x}}[/tex]
The outer tangent function has a derivative of:
-sec²(x).
Then the complete derivative is of:
[tex]\frac{dy}{dx} = -\sec^2{\left(\ln{x} - \frac{1}{\ln{x}}\right)} \times \left(\frac{1}{x} + \frac{1}{x\ln^2{x}}\right)[/tex]
At x = e, the numeric value of the derivative is given as follows:
[tex]\frac{dy}{dx}(e) = -\sec^2{0} \times \frac{2}{e}[/tex]
[tex]\frac{dy}{dx}(e) = -\frac{2}{e}[/tex]
More can be learned about partial derivatives at https://brainly.com/question/28728157
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