Have a maths assignment, which is proving a bit more challenging for me than I anticipated.
I'm having some trouble with a question that reads
Assume that $$ \sinh(y) = \frac{4sinh(x)-3}{4+3sinh(x)} $$
Show that $$ \frac{dy}{dx} = \frac{5}{4+3sinh(x)} $$
I started with implicit differentiation, and trying to get $$\frac{dy}{dx}$$ on its own, but I can't get to the answer. Maybe I made an error in my simplification? Or am I using the completely wrong method?
EDIT:
Retried, got to $$ \frac{dy}{dx}=\frac{25}{[4+3sinh(x)]^2} $$
Can I just take the square root of both numerator and denominator? Is there a different method I'm supposed to use?
Your lecturer is evil. Don't look if possible.
$\displaystyle \frac{d}{dx} \sinh(y) = \displaystyle \frac{d}{dx}\frac{4sinh(x)-3}{4+3sinh(x)}$
$\displaystyle cosh(y) \frac{dy}{dx} = {{25\,\cosh(x)}\over{\left(3\,\sinh( x)+4\right)^2}}$
$cosh(y) = \sqrt{1 + sinh^2(y)} = \displaystyle \frac{5 \, cosh(x)}{{4+3sinh(x)}}$
$\displaystyle \frac{5 \, cosh(x)}{{4+3sinh(x)}} \frac{dy}{dx} = {{25\,\cosh(x)}\over{\left(3\,\sinh(x)+4\right)^2}}$
$\displaystyle \frac{dy}{dx} = {{5}\over{4 + 3\,\sinh(x)}}$