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Chapter 3: Problem 61

Use the Chain Rule combined with other differentiation rules to find the derivative of the following functions. $$y=\left((x+2)\left(x^{2}+1\right)\right)^{4}$$

Short Answer

Expert verified
Question: Find the derivative of the function $y = [(x+2)(x^2+1)]^4$. Solution: The derivative of the given function is $\frac{dy}{dx} = 4((x+2)(x^2+1))^3 \cdot ((x^2+1) + (x+2)(2x))$.

Step by step solution

01

Apply the Chain Rule

Find the derivative of the outer function \(u(v) = v^4\) with respect to \(v\). We get: $$u'(v) = 4v^3$$ Find the derivative of the inner function \(v(x) = (x+2)(x^2+1)\) and denote it as \(v'(x)\). We will calculate this in the next step.
02

Use the product rule for the differentiation of the inner function

As \(v(x) = (x+2)(x^2+1)\), we need to apply the product rule for differentiation, which states that \((fg)' = f'g + fg'\). Let \(f(x) = x+2\) and \(g(x) = x^2+1\). The derivatives are: $$f'(x) = 1$$ $$g'(x) = 2x$$ Now apply the product rule: $$v'(x) = f'(x)g(x) + f(x)g'(x) = (1)(x^2+1) + (x+2)(2x)$$
03

Simplify the result

Now, we substitute \(u'(v)\) and \(v'(x)\) into the Chain Rule: $$\frac{dy}{dx} = u'(v(x)) \cdot v'(x) = 4v^3 \cdot v'(x) = 4((x+2)(x^2+1))^3 \cdot ((x^2+1) + (x+2)(2x))$$

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Most popular questions from this chapter

Cobb-Douglas production function The output of an economic system \(Q,\) subject to two inputs, such as labor \(L\) and capital \(K\) is often modeled by the Cobb- Douglas production function \(Q=c L^{a} K^{b} .\) When \(a+b=1,\) the case is called constant returns to scale. Suppose \(Q=1280, a=\frac{1}{3}, b=\frac{2}{3},\) and \(c=40\) a. Find the rate of change of capital with respect to labor, \(d K / d L\) b. Evaluate the derivative in part (a) with \(L=8\) and \(K=64\)

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Product Rule for three functions Assume that \(f, g,\) and \(h\) are differentiable at \(x\) a. Use the Product Rule (twice) to find a formula for \(\frac{d}{d x}(f(x) g(x) h(x))\) b. Use the formula in (a) to find \(\frac{d}{d x}\left(e^{2 x}(x-1)(x+3)\right)\)

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Proof of the Quotient Rule Let \(F=f / g\) be the quotient of two functions that are differentiable at \(x\) a. Use the definition of \(F^{\prime}\) to show that \(\frac{d}{d x}\left[\frac{f(x)}{g(x)}\right]=\lim _{h \rightarrow 0} \frac{f(x+h) g(x)-f(x) g(x+h)}{h g(x+h) g(x)}\) b. Now add \(-f(x) g(x)+f(x) g(x)\) (which equals 0) to the numerator in the preceding limit to obtain $$\lim _{h \rightarrow 0} \frac{f(x+h) g(x)-f(x) g(x)+f(x) g(x)-f(x) g(x+h)}{h g(x+h) g(x)}$$ Use this limit to obtain the Quotient Rule. c. Explain why \(F^{\prime}=(f / g)^{\prime}\) exists, whenever \(g(x) \neq 0\)
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