Question

Find the derivatives of the functions in Exercises 19–40.y = (1 / 18)(3x − 2)⁶ + (4 − (1 / 2x²))⁻¹

Accepted Answer

Identify the two main components of the function: the first term $$ \frac{1}{18}(3x - 2)^6 $$ and the second term $$ \left(4 - \frac{1}{2x^2}\right)^{-1} . We $$will differentiate each term separately and then combine the results. For the first term $$ \frac{1}{18}(3x - 2)^6 $$, apply the chain rule. The chain rule states that if you have a composite function $$ f(g(x)) $$, the derivative is $$ f'(g(x)) \cdot g'(x) . $$Here, let $$ u = 3x - 2 $$, so the derivative of $$ u^6  is$$ $$ 6u^5 $$, and the derivative of $$ u  is$$ $$ 3 . $$Therefore, the derivative of the first term is $$ \frac{1}{18} \cdot 6(3x - 2)^5 \cdot 3 . $$For the second term $$ \left(4 - \frac{1}{2x^2}\right)^{-1} $$, use the chain rule and the power rule. Let $$ v = 4 - \frac{1}{2x^2} $$, so the derivative of $$ v^{-1}  is$$ $$ -v^{-2} . $$The derivative of $$ v  is$$ $$ \frac{d}{dx}(4 - \frac{1}{2x^2}) = \frac{d}{dx}(4) - \frac{d}{dx}(\frac{1}{2x^2}) . $$The derivative of $$ 4  is$$ $$ 0 $$, and the derivative of $$ \frac{1}{2x^2}  is$$ $$ -\frac{1}{x^3} . $$Therefore, the derivative of $$ v  is$$ $$ \frac{1}{x^3} . $$Combine the derivatives of the two terms. The derivative of the first term is $$ \frac{1}{18} \cdot 18(3x - 2)^5 $$, and the derivative of the second term is $$ -\left(4 - \frac{1}{2x^2}\right)^{-2} \cdot \frac{1}{x^3} . $$Add the derivatives of the two terms to find the derivative of the entire function: $$ \frac{1}{18} \cdot 18(3x - 2)^5 + \left(-\left(4 - \frac{1}{2x^2}\right)^{-2} \cdot \frac{1}{x^3}\right) . $$Simplify the expression if necessary.

Find the derivatives of the functions in Exercises 19–40.

y = (1 / 18)(3x − 2)⁶ + (4 − (1 / 2x²))⁻¹

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Key Concepts

Power Rule

Chain Rule

Derivative of Reciprocal Function