91Ó°ÊÓ

We want to find the derivative of $$u = x$$ with respect to x. This is a simple power function, and its derivative is simply 1. So, $$\frac{du}{dx} = 1$$.

Now, we want to find the derivative of $$v = \tanh x$$ with respect to x. We know that $$\tanh x = \frac{\sinh x}{\cosh x}$$, where $$\sinh x = \frac{e^x - e^{-x}}{2}$$ and $$\cosh x = \frac{e^x + e^{-x}}{2}$$. We can use the quotient rule for derivatives, which states that for a function $$y = \frac{A(x)}{B(x)}$$, where A(x) and B(x) are differentiable functions of x, we have $$\frac{dy}{dx} = \frac{A'(x)B(x) - A(x)B'(x)}{(B(x))^2}$$. In our case, A(x) = sinh x, B(x) = cosh x. We first find the derivative of A(x) and B(x). For A'(x), we have $$\frac{d(\sinh x)}{dx} = \frac{d(\frac{e^x - e^{-x}}{2})}{dx} = \frac{e^x + e^{-x}}{2} = \cosh x$$. Similarly, for B'(x), we have $$\frac{d(\cosh x)}{dx} = \frac{d(\frac{e^x + e^{-x}}{2})}{dx} = \frac{e^x - e^{-x}}{2} = \sinh x$$. Now, using the quotient rule, we get the derivative of v(x) as: $$\frac{dv}{dx} = \frac{(\cosh x)(\cosh x) - (\sinh x)(\sinh x)}{(\cosh x)^2} = \frac{(\cosh x)^2 - (\sinh x)^2}{(\cosh x)^2} = 1 - \tanh^2 x = \text{sech}^2 x$$, where sech x is the hyperbolic secant function.

Now that we have the derivatives of u(x) and v(x), we can apply the product rule to find the derivative of y. Recall that $$\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$$. Plugging in the expressions for u, v, and their derivatives, we have: $$\frac{dy}{dx} = x\cdot\text{sech}^2 x + \tanh x\cdot1 = x\text{sech}^2 x + \tanh x$$ This is the derivative of the given function $$y=x \tanh x$$ with respect to x.