Re: These three derivative problems are mating in deriv.****
Use the info from your other post I commented on to do these.
The first one can be done by rewriting it as \(\displaystyle e^{2x}\cdot \sqrt{x^{2}+1}\cdot csc(x)\) and using the product rule for 3 terms from the other post.
The second one can be done using the chain rule.
The chain rule says \(\displaystyle \frac{dy}{dx}=\overbrace{\frac{dy}{du}}^{\text{outside}}\cdot \underbrace{\frac{du}{dx}}_{\text{inside}}\)
The chain rule is essentially the derivative of the inside times the derivative of the outside.
\(\displaystyle y=sec\left(\frac{x^{2}+1}{2x^{2}-1}\right)\)
Let \(\displaystyle u=\frac{x^{2}+1}{2x^{2}-1}\)
then we have \(\displaystyle sec(u)\)
Derivative of inside: \(\displaystyle u=\frac{x^{2}+1}{2x^{2}-1}=\frac{3}{2(2x^{2}-1)}+\frac{1}{2}\)
Using the quotient rule, the derivative is \(\displaystyle \frac{du}{dx}=\frac{-6x}{(2x^{2}-1)^{2}}\)
The derivative of the outside: \(\displaystyle \frac{dy}{du}=\frac{d}{du}[sec(u)]=sec(u)tan(u)du\)
Resub u and multiply by du/dx to finally get dy/dx. Then, algebraically simplify if possible.
If you are just learning derivatives, these can be rather challenging.
The third one can be done by using the formula derived from the second problem from the other post...the chain rule.
