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

Guest Mar 15, 2015

Best Answer 

 #1
avatar+92367 
+10

 -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)

 

  

CPhill  Mar 15, 2015
 #1
avatar+92367 
+10
Best Answer

 -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)

 

  

CPhill  Mar 15, 2015
 #2
avatar+94083 
+5

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)

Melody  Mar 16, 2015

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