what is the derivative of -6*sec(sin(5x^2+3x+2))

 -6*sec(sin(5x^2+3x+2)).....we will apply the -6 back at the end....using the chain rule several times we have

sec(sin(5x^2 + 3x + 2))*tan(sin(5x^2 + 3x + 2)) *cos(5x^2 + 3x + 2) * (10x + 3)

And applying the -6 back, we have

(-60x - 18)*sec(sin(5x^2 + 3x + 2))*tan(sin(5x^2 + 3x + 2)) *cos(5x^2 + 3x + 2)

Thanks CPhill, I just want to see if I can do it too.        (๑‵●‿●‵๑)

Mmm that looks tricky.

let

$$y= -6*sec(sin(5x^2+3x+2))$$

let

$$\\g = sin(5x^2+3x+2)\\\\ \frac{dg}{dx}=(10x+3)[cos(5x^2+3x+2)]\\\\\\ y=-6sec(g)\\\\ y=-6(cos(g))^{-1}\\\\ \frac{dy}{dg}=6(cos(g))^{-2}(-sin(g))\\\\ \frac{dy}{dg}=\frac{-6sin(g)}{cos^2(g)}\\\\\\ \frac{dy}{dx}=\frac{dy}{dg}\times \frac{dg}{dx}\\\\ \frac{dy}{dx}=\frac{-6sin(g)}{cos^2(g)}\times (10x+3)[cos(5x^2+3x+2)]\\\\$$

$$\\\frac{dy}{dx}=\frac{-6sin(g)(10x+3)[cos(5x^2+3x+2)]}{cos^2(g)}\\\\ \frac{dy}{dx}=\frac{-6sin(sin(5x^2+3x+2))(10x+3)[cos(5x^2+3x+2)]}{cos^2(sin(5x^2+3x+2))}\\\\ \frac{dy}{dx}=-6tan(sin(5x^2+3x+2)sec(sin(5x^2+3x+2))(10x+3)[cos(5x^2+3x+2)]}\\\\ \frac{dy}{dx}=-(60x+18)tan(sin(5x^2+3x+2)sec(sin(5x^2+3x+2))[cos(5x^2+3x+2)]}\\\\$$

WOW this is the same as CPhill's answer     (๑‵●‿●‵๑)