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Find the derivatives of the given functions. Assume that \(a, b, c,\) and \(k\) are constants. $$y=\frac{1}{r^{7 / 2}}$$

Short Answer

Expert verified
The derivative is \( \frac{dy}{dr} = -\frac{7}{2r^{9/2}} \).

Step by step solution

01

Rewrite the Function Using Negative Powers

The given function is \[ y = \frac{1}{r^{7/2}} \]To simplify differentiation, rewrite this using negative exponents:\[ y = r^{-7/2} \]
02

Apply the Power Rule for Derivatives

The power rule states that the derivative of \( x^n \) is \( n \cdot x^{n-1} \). Apply this rule to \( y = r^{-7/2} \):\[ \frac{dy}{dr} = \left(-\frac{7}{2}\right) \cdot r^{- rac{7}{2} - 1} \] which simplifies to:\[ \frac{dy}{dr} = -\frac{7}{2} \cdot r^{-\frac{9}{2}} \]
03

Simplify the Derivative

Simplify the expression for clarity by converting the negative exponent to a fraction:\[ \frac{dy}{dr} = -\frac{7}{2} \cdot \frac{1}{r^{9/2}} \] This can be rewritten as:\[ \frac{dy}{dr} = -\frac{7}{2r^{9/2}} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Power Rule
The power rule is a handy tool when differentiating terms that are powers of a variable. In calculus, the power rule states that if you have a function in the form of \( x^n \), the derivative of that function is \( n \cdot x^{n-1} \). This rule drastically simplifies the process of finding derivatives, allowing us to quickly compute them without more complex algebraic manipulations.
For example, when we applied the power rule to the function \( y = r^{-7/2} \), which we had rewritten using negative exponents, we simply multiplied the exponent \(-7/2\) by the function \( r \) raised to the new power \(-7/2 - 1 \). This simple multiplication and subtraction provided us the derivative's structure: \( \frac{dy}{dr} = -\frac{7}{2} \cdot r^{-9/2} \).
Negative Exponents
Negative exponents can seem intimidating, but they offer a convenient way to express more complex divisions involving powers. When you see a function with a term in the denominator, like \( \frac{1}{r^{7/2}} \), you can rewrite it using a negative exponent as \( r^{-7/2} \). This recasting is crucial when differentiating, as it aligns with the power rule's format.
By understanding that \( r^{-n} = \frac{1}{r^n} \), you simplify the expression and make differentiation straightforward. This conversion quickly sets the stage for applying other calculus rules, ensuring that you can easily handle intricate expressions by treating them with standard calculus techniques, like the power rule.
Simplification of Derivatives
Simplification is often the final step after applying differentiation rules, vital for making derivatives easier to understand and use. After finding the derivative, you might encounter complex expressions often involving negative exponents.
When you simplify \( -\frac{7}{2} \cdot r^{-9/2} \), you transform it into \( -\frac{7}{2} \cdot \frac{1}{r^{9/2}} \). This further reduces to \( -\frac{7}{2r^{9/2}} \), a clearer form that can be more readily used in further calculations or practical applications. By converting the expression with negative exponents back to fractions, you make the derivative more accessible and intuitive, especially in function analysis or solving more extensive mathematical problems.

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Most popular questions from this chapter

A wave travels along a string that is joined to a thicker rope. The wave both reflects back along the string and is transmitted to the rope. For positive constants \(k_{1}, k_{2}, w\) and time \(t,\) the wave along the string, given by \(x<0,\) is $$f(x)=\cos \left(k_{1} x-w t\right)+R \cos \left(-k_{1} x-w t\right)$$and the wave along the rope, given by \(x>0,\) is $$ g(x)=T \cos \left(k_{2} x-w t\right)$$ For every value of \(t,\) the two waves have the same tangent line at \(x=0,\) so they have the same value and same slope at \(x=0 .\) Use this fact to show that $$R=\frac{k_{1}-k_{2}}{k_{1}+k_{2}} \quad \text { and } \quad T=\frac{2 k_{1}}{k_{1}+k_{2}}$$ These amplitudes are called the reflection coefficient, \(R\) and the transmission coefficient, \(T\).

Explain what is wrong with the statement. $$\cosh ^{2} x+\sinh ^{2} x=1$$

Give an explanation for your answer. The function \(\tanh x\) is odd, that is, \(\tanh (-x)=\) \(-\tanh x\)

Let \(f\) be a differentiable function and let \(L\), be the linear function \(L(x)=f(a)+k(x-a)\) for some constant \(a\) Decide whether the following statements are true or false for all constants \(k .\) Explain your answer. (a) \(L\) is the local linearization for \(f\) near \(x=a\) (b) If \(\lim _{x \rightarrow a}(f(x)-L(x))=0,\) then \(L\) is the local linearization for \(f\) near \(x=a\).

Explain what is wrong with the statement. The linear approximation for \(F(x)=x^{3}+1\) near \(x=0\) is an underestimate for the function \(F\) for all \(x, x \neq 0\)

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