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PTA

PTA. “Power, Trig, Argument” Used when you have a function raised to a power. By Olivia Root. Starting problem: Find the derivative of y=5sin³(x²)= (5sin(x²)) ³

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PTA

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  1. PTA “Power, Trig, Argument” Used when you have a function raised to a power. By Olivia Root

  2. Starting problem: Find the derivative of y=5sin³(x²)= (5sin(x²)) ³ • First, RECOGNIZE that this is a PTA problem. There is a trig function that is all raised to a power, not just a power in the argument (i.e. x²) • Next, just think PTA…

  3. POWER y=5sin(x²)³ Bring the power down in front of the function as you would with any derivative, and replace it with an exponent that is one less than the original, so: y=(3)5sin²(x²)

  4. TRIG y=(3)5sin²(x²) Next is the trig step. In this step, you will add the derivate of your trig function to your answer. In this case, the derivative of sine is…cosine! So you multiply what you already have by cosine of the argument, as follows: y=(3)5sin²(x²)cos(x²)

  5. ARGUMENT y=(3)5sin²(x²)cos(x²) Almost there! Here is our last step of the PTA process. Multiply your answer by the derivative of your original argument, x². As we all know, this is 2x. So: y=(3)5sin²(x²)cos(x²)(2x)

  6. And… Clean it up! y=(3)5sin²(x²)cos(x²)(2x) = 15sin²(x²)cos(x²)(2x) = (30x)sin²(x²)cos(x²) YAY!

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