What is a Derivative?
The 5 Main Derivative Rules
Special Case Derivatives: Your new multiplication tables
1 of 2

Example 2: Inverse Trig Derivatives

Differentiate f ( x ) = arcsec ( 3 x 2 8 x + 5 ) with respect to x.

Step 1: Simplify and look for algebraic rewrites.

Here the equation is a simplified as we need it, and there are no algebra rewrites.

You always have to check though.

f ( x ) = arcsec ( 3 x 2 8 x + 5 )

Step 2: Identify your equation as the one of the six inverse trig function special cases.

Here the equation is the arcsecant or inverse secant function, which is one of our six inverse trig function special cases, and it does have more than just an x inside it.

f ( x ) = arcsec ( 3 x 2 8 x + 5 )

Step 3 (If it has more than just an x ): Run the standard chain rule process with whatever replaced your basic x as the inside piece.

Break the problem up into two these two bite-size problems.

IN = SOMETHING and OUT = SOMETHING

IN = 3 x 2 8 x + 5 OUT = arcsec ( 3 x 2 8 x + 5 )

Take the derivatives of those two bite-size problems to find your DIN and your DOUT .

DIN : The derivative of IN only requires you to apply the power rule.

Bring the power down.

Subtract 1 from the power.

DOUT : When you are taking the derivative of the OUT piece, you ignore what is inside the parenthesis, and perform the derivative rule only for the outside piece, and leave the IN alone.

With the six inverse trig function special cases, you will plug the IN piece everywhere you see an x in the inverse trig derivative, f ( x ) .

f ( x ) = 1 | x | x 2 1

IN = 3 x 2 8 x + 5 OUT = arcsec ( 3 x 2 8 x + 5 )

DIN = 2 3 x 2 1 1 8 x 1 1 + 0 DOUT = 1 | 3 x 2 8 x + 5 | ( 3 x 2 8 x + 5 ) 2 1

DIN = 6 x 1 8 x 0 + 0

DIN = 6 x 8

Bring it all back together following the chain rule recipe: (DIN)(DOUT).

f ( x ) = ( DIN ) ( DOUT )

f ( x ) = ( 6 x 8 ) ( 1 | 3 x 2 8 x + 5 | ( 3 x 2 8 x + 5 ) 2 1 )

Final Result:

The derivative of f ( x ) = arc sec ( 3 x 2 8 x + 5 ) is f ( x ) = ( 6 x 8 ) ( 1 | 3 x 2 8 x + 5 | ( 3 x 2 8 x + 5 ) 2 1 ) .

Most of the time you will want to leave the derivative as it is and not try to expand it or simplify it.

My rule of thumb is if someone wants the derivative give them this unless they ask you to simplify further.

If you need to do more work, like finding a second derivative, you would want to consider expanding and simplifying to make your life easier.

Meaning:

The equation for finding the slope of any tangent line at any x-value of f ( x ) = arc sec ( 3 x 2 8 x + 5 ) is f ( x ) = ( 6 x 8 ) ( 1 | 3 x 2 8 x + 5 | ( 3 x 2 8 x + 5 ) 2 1 ) .

The instantaneous rate of change for every x-value of f ( x ) = arc sec ( 3 x 2 8 x + 5 ) , is found by using the derivative equation, f ( x ) = ( 6 x 8 ) ( 1 | 3 x 2 8 x + 5 | ( 3 x 2 8 x + 5 ) 2 1 ) .

The slope of f ( x ) = arc sec ( 3 x 2 8 x + 5 ) at any single x-value can be found by plugging it into the derivative, f ( x ) = ( 6 x 8 ) ( 1 | 3 x 2 8 x + 5 | ( 3 x 2 8 x + 5 ) 2 1 ) .

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