Derivatives of Trig functions: y= sin(tan(1+x^3)

[maeveoneill]

Just want some help to see if I'm doing my derivatives right.

1) y= sin(tan(1+x^3)

y''= cos(tan(1+ x^3))(sec^2x(1+x^3)(3x^2)

can someone just tell me if this is right and if im on the right track.. and offer how i would simplify it??

2) 2e^xy = x + y

2e^xy(x) dy/dx = 1 + dy/dx
dy/dx (2e^xy -1) =1
dy/dx = 1/ 2xe^xy -1

Thank you!!

 

[soroban]

Hello, maeveoneill!

You're doing great!

\(\displaystyle 1)\; y\:= \:\sin\left[\tan\left(1\,+\,x^3\right)\right]\)

\(\displaystyle y' \:= \:\cos\left[\tan\left(1\,+\,x^3\right)\right]\,\cdot\,\sec^2\left(1\,+\,x^3\right)\,\cdot\,(3x^2)\)

It can't be simplified.

\(\displaystyle 2)\;2e^{xy}\:=\:x\,+\,y\)

\(\displaystyle 2e^{xy}(x)\frac{dy}{dx} \:=\: 1\,+\,\frac{dy}{dx}\)

\(\displaystyle \frac{dy}{dx}\left(2xe^{xy}\,-\,1) \:=\:1\)

\(\displaystyle \frac{dy}{dx} \:=\: \frac{1}{2xe^{xy}\,-\,1}\)

Nice work!