<p>could someone help me differentiate this problem:
y=[(e^x)-(e^-x)]/[(e^x)+(e^-x)]</p>
<p>How do you even do this??? I'm missing something cause this should be a cake problem but idk what.</p>
<p>I worked it out quickly and got:
y’ = - [(e^x - e^-x)^2] / [(e^x + e^-x)^2]</p>
<p>Do you have the answer to this, so I may double check if I did it correctly?</p>
<p>But here is my working, I’m not sure of your level in calculus so I tried to go through the steps I used exactly.</p>
<p>y’ = [(u’v) - (v’u)] / v^2 (quotient rule)</p>
<p>u = (e^x)-(e^-x)
u’ = (e^x)-(-e^-x)
= (e^x + e^-x)</p>
<p>v = (e^x)+(e^-x)
v’= (e^x) + (-e^-x)
= (e^x - e^-x)</p>
<p>(e^u differentiated is (e^u)(u’))</p>
<p>y’ = [(e^x + e^-x)(e^x + e^-x)] - [(e^x - e^-x)(e^x - e^-x)] / [(e^x + e^-x)^2]</p>
<p>= [(e^x + e^-x)^2] - [(e^x - e^-x)^2] / [(e^x + e^-x)^2]</p>
<p>(now to separate them)</p>
<p>= [(e^x + e^-x)^2] / [(e^x + e^-x)^2] - [(e^x - e^-x)^2] / [(e^x + e^-x)^2]</p>
<p>(the first cancels out and leaves us with)</p>
<p>= - [(e^x - e^-x)^2] / [(e^x + e^-x)^2]</p>
<p>To reiterate I did this quickly, please check my work and if possible I’d appreciate if anyone else could take a look.</p>
<p>You can try this:
y=(e^x - e^-x)/(e^x + e^-x) = 1 - (2e^-x)/(e^x+e^-x)</p>
<p>Differentiate:
dy/dx= -d[(2e^-x)/(e^x + e^-x)]/dx</p>
<p>Let’s deem (e^x + e^-x) as v.</p>
<p>= [(e^x + e^-x)(-2e^-x) - (2e^-x)(e^x - e^-x)]/v^2</p>
<p>= [(-2e^-x)(e^x + e^-x + e^x - e^-x)]/v^2</p>
<p>= -4(e^-x)(e^x)/v^2</p>
<p>= -4/v^2</p>
<p>And the rest is self-explanatory.</p>
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<p>Isn’t this tanh(x)?</p>
<p>sech(x)sech(x)
lollllll</p>