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If f(x)=\frac{(3x^{2}-4)^{5}}{2x-1},then find {f}'(x).

Question:

If {eq}\displaystyle f(x) = \frac{(3x^{2} - 4)^{5}}{2x - 1}, {/eq} then find {eq}{f}'(x). {/eq}

Differentiation:

Quotient Rule :

When question is about differentiating function dividing existing function,

Or in other words fraction of functions is given then we need to apply quotient rule.

Formula,

{eq}(\frac{x}{y})' = \frac{xy'+x'y}{y^2} {/eq}

Answer and Explanation:

We have,

{eq}\displaystyle f(x) = \frac{(3x^{2} - 4)^{5}}{2x - 1}, {/eq}

Now,

Differentiating both sides,

{eq}\displaystyle f'(x) = (\frac{(3x^{2} - 4)^{5}}{2x - 1})' \\ \displaystyle f'(x) = \frac{((3x^{2} - 4)^{5})'(2x-1) - (3x^{2} - 4)^{5}(2x-1)'}{(2x - 1)^2} \\ \displaystyle f'(x) = \frac{(5(3x^{2} - 4)^{4}(6x))(2x-1) - 2(3x^{2} - 4)^{5}}{(2x - 1)^2} \\ \displaystyle f'(x) = \frac{(30x(2x-1) - 2(3x^{2} - 4))(3x^{2} - 4)^4}{(2x - 1)^2} \\ \displaystyle f'(x) = \frac{(60x^2- 30x - 6x^{2} + 8)(3x^{2} - 4)^4}{(2x - 1)^2} \\ {/eq}

{eq}\therefore \color{blue}{\displaystyle f'(x) = \frac{(60x^2- 30x - 6x^{2} + 8)(3x^{2} - 4)^4}{(2x - 1)^2} } {/eq}


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Basic Calculus: Rules & Formulas

from Calculus: Tutoring Solution

Chapter 3 / Lesson 6
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