Determine $\frac{dy}{dx}$ for the following y, where: a) $y = \sqrt{x}$ b) $y = \frac{2}{x^3}$ c) $y = \sqrt[5]{x}$ d) $y = \frac{5}{x^6}$ e) $y = x\sqrt{x}$ f) $y = 2x^3\sqrt{x}$
Understand the Problem
The question asks to determine the derivative, denoted as $\frac{dy}{dx}$, for each of the given functions of $y$. This involves applying differentiation rules to find the rate of change of $y$ with respect to $x$ for each given $y(x)$. We need to apply rules such as power rule, product rule and quotient rule of differentiation.
Answer
a) $\frac{dy}{dx} = \frac{1}{2\sqrt{x}}$ b) $\frac{dy}{dx} = -\frac{6}{x^4}$ c) $\frac{dy}{dx} = \frac{1}{5\sqrt[5]{x^4}}$ d) $\frac{dy}{dx} = -\frac{30}{x^7}$ e) $\frac{dy}{dx} = \frac{3}{2}\sqrt{x}$ f) $\frac{dy}{dx} = 7x^2\sqrt{x}$
Answer for screen readers
a) $\frac{dy}{dx} = \frac{1}{2\sqrt{x}}$ b) $\frac{dy}{dx} = -\frac{6}{x^4}$ c) $\frac{dy}{dx} = \frac{1}{5\sqrt[5]{x^4}}$ d) $\frac{dy}{dx} = -\frac{30}{x^7}$ e) $\frac{dy}{dx} = \frac{3}{2}\sqrt{x}$ f) $\frac{dy}{dx} = 7x^2\sqrt{x}$
Steps to Solve
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Derivative of $y = \sqrt{x}$ Rewrite $y = \sqrt{x}$ as $y = x^{\frac{1}{2}}$. Apply the power rule: $\frac{dy}{dx} = \frac{1}{2}x^{\frac{1}{2} - 1} = \frac{1}{2}x^{-\frac{1}{2}} = \frac{1}{2\sqrt{x}}$.
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Derivative of $y = \frac{2}{x^3}$ Rewrite $y = \frac{2}{x^3}$ as $y = 2x^{-3}$. Apply the power rule: $\frac{dy}{dx} = 2(-3)x^{-3 - 1} = -6x^{-4} = -\frac{6}{x^4}$.
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Derivative of $y = \sqrt[5]{x}$ Rewrite $y = \sqrt[5]{x}$ as $y = x^{\frac{1}{5}}$. Apply the power rule: $\frac{dy}{dx} = \frac{1}{5}x^{\frac{1}{5} - 1} = \frac{1}{5}x^{-\frac{4}{5}} = \frac{1}{5\sqrt[5]{x^4}}$.
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Derivative of $y = \frac{5}{x^6}$ Rewrite $y = \frac{5}{x^6}$ as $y = 5x^{-6}$. Apply the power rule: $\frac{dy}{dx} = 5(-6)x^{-6 - 1} = -30x^{-7} = -\frac{30}{x^7}$.
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Derivative of $y = x\sqrt{x}$ Rewrite $y = x\sqrt{x}$ as $y = x \cdot x^{\frac{1}{2}} = x^{\frac{3}{2}}$. Apply the power rule: $\frac{dy}{dx} = \frac{3}{2}x^{\frac{3}{2} - 1} = \frac{3}{2}x^{\frac{1}{2}} = \frac{3}{2}\sqrt{x}$.
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Derivative of $y = 2x^3\sqrt{x}$ Rewrite $y = 2x^3\sqrt{x}$ as $y = 2x^3 \cdot x^{\frac{1}{2}} = 2x^{\frac{7}{2}}$. Apply the power rule: $\frac{dy}{dx} = 2(\frac{7}{2})x^{\frac{7}{2} - 1} = 7x^{\frac{5}{2}} = 7x^2\sqrt{x}$.
a) $\frac{dy}{dx} = \frac{1}{2\sqrt{x}}$ b) $\frac{dy}{dx} = -\frac{6}{x^4}$ c) $\frac{dy}{dx} = \frac{1}{5\sqrt[5]{x^4}}$ d) $\frac{dy}{dx} = -\frac{30}{x^7}$ e) $\frac{dy}{dx} = \frac{3}{2}\sqrt{x}$ f) $\frac{dy}{dx} = 7x^2\sqrt{x}$
More Information
The power rule of differentiation states that if $y = ax^n$, where $a$ is a constant and $n$ is a real number, then $\frac{dy}{dx} = nax^{n-1}$. This rule has been applied to solve all the sub-questions.
Tips
A common mistake is forgetting to rewrite the functions in the form $ax^n$ before applying the power rule. For example, not rewriting $\sqrt{x}$ as $x^{\frac{1}{2}}$ or $\frac{1}{x^3}$ as $x^{-3}$ before differentiating. Another common mistake is to make errors in the arithmetic when calculating the new exponent after applying the power rule (i.e., $n-1$).
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