9.4 First Derivatives of the Quotient of Two Polynomials Posted on April 22, 2020 by user Find the Derivatives of a Quotient using Quotient Rule Method 1: The Quotient Rule Example: Method 2: (Differentiate Directly) Example: Given that y = x 2 2 x + 1 , find d y d x Solution: y = x 2 2 x + 1 d y d x = ( 2 x + 1 ) ( 2 x ) − x 2 ( 2 ) ( 2 x + 1 ) 2 = 4 x 2 + 2 x − 2 x 2 ( 2 x + 1 ) 2 = 2 x 2 + 2 x ( 2 x + 1 ) 2 Practice 1: Given that y = 4 x 3 ( 5 x + 1 ) 3 , find d y d x Solution: y = 4 x 3 ( 5 x + 1 ) 3 d y d x = ( 5 x + 1 ) 3 ( 12 x 2 ) − 4 x 3 .3 ( 5 x + 1 ) 2 .5 [ ( 5 x + 1 ) 3 ] 2 = ( 5 x + 1 ) 3 ( 12 x 2 ) − 60 x 3 ( 5 x + 1 ) 2 ( 5 x + 1 ) 6 = ( 12 x 2 ) ( 5 x + 1 ) 2 [ ( 5 x + 1 ) − 5 x ] ( 5 x + 1 ) 6 = ( 12 x 2 ) ( 5 x + 1 ) 2 ( 1 ) ( 5 x + 1 ) 6 = 12 x 2 ( 5 x + 1 ) 4
9.4 First Derivatives of the Quotient of Two Polynomials Posted on April 22, 2020 by user 9.4 Find the Derivatives of a Quotient using Quotient Rule Method 1 The Quotient Rule Example: Method 2 (Differentiate Directly) Example: Given that y = x 2 2 x + 1 , find d y d x Solution: y = x 2 2 x + 1 d y d x = ( 2 x + 1 ) ( 2 x ) − x 2 ( 2 ) ( 2 x + 1 ) 2 = 4 x 2 + 2 x − 2 x 2 ( 2 x + 1 ) 2 = 2 x 2 + 2 x ( 2 x + 1 ) 2 Practice 1: Given that y = 4 x 3 ( 5 x + 1 ) 3 , find d y d x Solution: y = 4 x 3 ( 5 x + 1 ) 3 d y d x = ( 5 x + 1 ) 3 ( 12 x 2 ) − 4 x 3 .3 ( 5 x + 1 ) 2 .5 [ ( 5 x + 1 ) 3 ] 2 = ( 5 x + 1 ) 3 ( 12 x 2 ) − 60 x 3 ( 5 x + 1 ) 2 ( 5 x + 1 ) 6 = ( 12 x 2 ) ( 5 x + 1 ) 2 [ ( 5 x + 1 ) − 5 x ] ( 5 x + 1 ) 6 = ( 12 x 2 ) ( 5 x + 1 ) 2 ( 1 ) ( 5 x + 1 ) 6 = 12 x 2 ( 5 x + 1 ) 4
9.2 First Derivative for Polynomial Function Posted on April 22, 2020 by user 9.2 First Derivative for Polynomial Function (A) Differentiating a Constant (B) Differentiating Variable with Index n (C) Differentiating a Linear Function (D) Differentiating a Polynomial Function (E) Differentiating Fractional Function (F) Differentiating Square Root Function Posts navigation Newer posts →