Find the Derivative:

Question

This problem for my Calculus 1 class has got me stumped. I am not sure on where to start for this problem. Any help would be much appreciated.

$y=x\tanh^{-1}(x) + \ln(\sqrt{1-x^2})$

Answer 1

HINTS:

$$\text{arctanh}(x)=\frac12\log\left(\frac{1+x}{1-x}\right)$$

Then, use the product rule along with the chain rule.

Can you proceed now?

Answer 2

$${ y }^{ \prime }={ \left( x\tanh ^{ -1 } (x)+\ln { \left( \sqrt { 1-x^{ 2 } } \right) } \right) }^{ \prime }={ \left( x\tanh ^{ -1 } (x) \right) }^{ \prime }+{ \left( \ln { \left( \sqrt { 1-x^{ 2 } } \right) } \right) }^{ \prime }=\\ ={ x }^{ \prime }\tanh ^{ -1 } (x)+x{ \left( \tanh ^{ -1 } (x) \right) }^{ \prime }+\frac { 1 }{ \sqrt { 1-x^{ 2 } } } { \left( \sqrt { 1-x^{ 2 } } \right) }^{ \prime }=\\ =\tanh ^{ -1 } (x)+\frac { x }{ 1-{ x }^{ 2 } } -\frac { 2x }{ 1-{ x }^{ 2 } } =\tanh ^{ -1 } (x)-\frac { x }{ 1-{ x }^{ 2 } } $$

Answer 3

$y=x\tanh^{−1}(x)+\ln(\sqrt{1−x^2})$

differentiation is linear $\frac {dy}{dx} = \frac d{dx} (x\tanh^{−1}(x)) +\frac d{dx}(\ln(\sqrt{1−x^2})$

next we need to know the product rule and the chain rule

$\frac {dy}{dx} = \tanh^{−1}(x) + x\frac d{dx} \tanh^{−1}(x) +\frac {1}{\sqrt{1−x^2}} \frac d{dx}\sqrt{1−x^2}\\ \frac {dy}{dx} = \tanh^{−1}(x) + x\frac d{dx} \tanh^{−1}(x) +\frac {1}{\sqrt{1−x^2}} \frac {-x}{\sqrt{1−x^2}}\\ \frac {dy}{dx} = \tanh^{−1}(x) + x\frac d{dx} \tanh^{−1}(x) -\frac {x}{1−x^2}$

Which leave one nasty left to tackle

$u = \tanh^{-1} x\\ \tanh u = x\\ \sech^2 u \frac {du}{dx} = 1\\ \frac {du}{dx} = \frac 1{\sech^2 u}\\ \frac {du}{dx} = \frac 1{1-\tanh^2 u}\\ \frac {du}{dx} = \frac 1{1- x^2}\\ $

$\frac {dy}{dx} = \tanh^{−1}(x) + \frac {x}{1-x^2} -\frac {x}{1−x^2}\\ \frac {dy}{dx} = \tanh^{−1}(x)$

Answer 4

First note that for $f(x)=\tanh(x)$ , using the quotient rule from the definition of $\tanh x =\frac{\sinh x}{\cosh x}$, we can write the derivative as: $$ f'(x)=1-\tanh^2 x $$ Now using the derivative of the inverse function: $$ \frac {d}{dx}f^{-1}(x)=\frac{1}{f'(f^{-1}(x))} $$ we have:

$$ \frac{d}{dx}\tanh^{-1}(x)=\frac{1}{1-x^2} $$

so your derivative, using product and chain rules, becomes: $$ y'=\tanh^{-1}(x)+\frac{x}{1-x^2}+\frac{1}{\sqrt{1-x^2}}\cdot\left(\frac{-2x}{2\sqrt{1-x^2}} \right)=\tanh^{-1}(x)+\frac{x}{1-x^2}-\frac{x}{1-x^2}=\tanh^{-1}(x) $$