# Find the Derivative:

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=xtanh−1(x)+ln(√1−x2)y=x\tanh^{-1}(x) + \ln(\sqrt{1-x^2})

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Product and chain rule allow you to deal with all the terms up to the derivative of tanh−1\tanh^{-1}. For that one you recall that f∘f−1(x)=xf\circ f^{-1}(x)=x and use the chain rule to get an expression for (f−1)′(f^{-1})’ (ultimately you obtain the first expression in this page en.m.wikipedia.org/wiki/Inverse_functions_and_differentiation )
– b00n heT
2 days ago

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4

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HINTS:

arctanh(x)=12log(1+x1−x)\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?

Yes I think so. I will comment if I still don’t get it. Thank you so much.
– J. Armstrong
2 days ago

You’re welcome. My pleasure!
– Dr. MV
2 days ago

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– Dr. MV
2 days ago

y′=(xtanh−1(x)+ln(√1−x2))′=(xtanh−1(x))′+(ln(√1−x2))′==x′tanh−1(x)+x(tanh−1(x))′+1√1−x2(√1−x2)′==tanh−1(x)+x1−x2−2×1−x2=tanh−1(x)−x1−x2{ 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 } }

y=xtanhâˆ′1(x)+ln(√1âˆ′x2)y=x\tanh^{âˆ’1}(x)+\ln(\sqrt{1âˆ’x^2})

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

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

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

First note that for f(x)=\tanh(x)f(x)=\tanh(x) , using the quotient rule from the definition of \tanh x =\frac{\sinh x}{\cosh x}\tanh x =\frac{\sinh x}{\cosh x}, we can write the derivative as:

f'(x)=1-\tanh^2 x

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))}

\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}

\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)

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)