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Chapter 16: Problem 1

Differentiate each of the following functions: (a) \(\left(x^{3}+2\right)^{6}\) (b) \(\sqrt{\sin x}\) (c) \(\left(\mathrm{e}^{x}+1\right)^{7}\) (d) \((\cos 2 x)^{5}\) (e) \(\ln (x+1)\)

Short Answer

Expert verified
Question: Differentiate the following functions: (a) \((x^{3}+2)^{6}\) (b) \(\sqrt{\sin x}\) (c) \((e^{x}+1)^{7}\) (d) \((\cos 2x)^5\) (e) \(\ln(x+1)\) Answer: (a) \((x^{3}+2)^{6}' = 18x^2(x^3+2)^5\) (b) \((\sqrt{\sin x})' = \frac{\cos x}{2\sqrt{\sin x}}\) (c) \(((e^x+1)^7)' = 7(e^x+1)^6 \cdot e^x\) (d) \((\cos 2x)^5' = -10(\cos 2x)^4\sin 2x\) (e) \((\ln(x+1))' = \frac{1}{x+1}\)

Step by step solution

01

(a) Differentiating \((x^{3}+2)^{6}\)

To differentiate this function, we will use the chain rule. The chain rule states that if we have a function of the form \((f(g(x)))'\), then the derivative is given by: \((f(g(x)))' = f'(g(x)) \cdot g'(x)\) Let \(f(x) = x^6\) and \(g(x) = x^3 + 2\). Then our function becomes \(f(g(x)) = (x^3 + 2)^6\). Step 1: Differentiate \(f(x)\) and \(g(x)\): \(f'(x) = 6x^5\) \(g'(x) = 3x^2\) Step 2: Apply the chain rule: \((f(g(x)))' = f'(g(x)) \cdot g'(x)\) \((x^3 + 2)^6)' = 6(x^3 + 2)^5 \cdot 3x^2\) Step 3: Simplify the result: \((x^3 + 2)^6)' = 18x^2(x^3 + 2)^5\)
02

(b) Differentiating \(\sqrt{\sin x}\)

To differentiate this function, we'll use the chain rule again. We'll rewrite the function as \((\sin x)^{1/2}\), and proceed with the differentiation. Step 1: Rewrite the function: \(\sqrt{\sin x} = (\sin x)^{1/2}\) Step 2: Differentiate \(f(x) = x^{1/2}\) and \(g(x) = \sin x\): \(f'(x) = \frac{1}{2}x^{-1/2}\) \(g'(x) = \cos x\) Step 3: Apply the chain rule: \((f(g(x)))' = f'(g(x)) \cdot g'(x)\) \(((\sin x)^{1/2})' = \frac{1}{2}(\sin x)^{-1/2} \cdot \cos x\) Step 4: Simplify the result: \(((\sin x)^{1/2})' = \frac{\cos x}{2\sqrt{\sin x}}\)
03

(c) Differentiating \((e^{x}+1)^{7}\)

Here, we apply the chain rule once again by differentiating \(f(x) = x^7\) and \(g(x) = e^x + 1\). Step 1: Differentiate \(f(x)\) and \(g(x)\): \(f'(x) = 7x^6\) \(g'(x) = e^x\) Step 2: Apply the chain rule: \((f(g(x)))' = f'(g(x)) \cdot g'(x)\) \(((e^x + 1)^7)' = 7(e^x + 1)^6 \cdot e^x\)
04

(d) Differentiating \((\cos 2x)^5\)

Let \(f(x) = x^5\) and \(g(x) = \cos 2x\). Apply the chain rule and trigonometric differentiation rule. Step 1: Differentiate \(f(x)\) and \(g(x)\): \(f'(x) = 5x^4\) \(g'(x) = -2\sin 2x\) Step 2: Apply the chain rule: \((f(g(x)))' = f'(g(x)) \cdot g'(x)\) \(((\cos 2x)^5)' = 5(\cos 2x)^4 \cdot (-2\sin 2x)\) Step 3: Simplify the result: \(((\cos 2x)^5)' = -10(\cos 2x)^4\sin 2x\)
05

(e) Differentiating \(\ln(x+1)\)

Differentiate the natural logarithm function using the chain rule. Step 1: Differentiate \(f(x) = \ln x\) and \(g(x) = x+1\): \(f'(x) = \frac{1}{x}\) \(g'(x) = 1\) Step 2: Apply the chain rule: \((f(g(x)))' = f'(g(x)) \cdot g'(x)\) \((\ln(x + 1))' = \frac{1}{x + 1} \cdot 1\) Step 3: Simplify the result: \((\ln(x + 1))' = \frac{1}{x + 1}\)

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Most popular questions from this chapter

Use parametric differentiation to find \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) given (a) \(x=1+t, y=2+3 t+t^{2}\) (b) \(x=\sin t, y=\cos t\) (c) \(x=t^{2}, y=t^{3}\) (d) \(x=\mathrm{e}^{t}, y=\mathrm{e}^{t}+t\) (e) \(x=\sqrt{t}, y=1+\ln t\)

State the range of values of \(x\) for which each of the following functions is (i) concave up (ii) concave down: (a) \(y=\frac{x^{3}}{6}-\frac{5 x^{2}}{2}+3 x-9\) (b) \(y=3+x+\frac{x^{2}}{2}-\frac{x^{4}}{12}\) (c) \(y=\mathrm{e}^{x}-100 x-100\) (d) \(y=(x-1)^{4}\) (e) \(y=\frac{x^{2}}{2} \ln x-\frac{x^{2}}{4}\)

Find \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) where \(y\) is given by (a) \(x^{3} \mathrm{e}^{2 x}\) (b) \(-3 \sin 2 x \cos 5 x\) (c) \(\left(x^{2}+1\right) \mathrm{e}^{-x}\) (d) \(\sin x \sin 2 x \sin 3 x\) (e) \(x \tan 3 x\).

Determine the location of all maximum and minimum points of the following functions: (a) \(y=x^{2}-4 x\) (b) \(y=x^{2}-5 x+4\) (c) \(y=10+3 x-x^{2}\) (d) \(y=\frac{x^{3}}{3}-\frac{x^{2}}{2}+1\) (e) \(y=x^{3}-27 x\)

Use logarithmic differentiation to differentiate $$ y=x^{x} $$
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