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Calculus 1 : Differential Functions

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Example Questions

Example Question #81 : Other Differential Functions

Find the derivative of

$f(x) = \sin(x)\cos(2x)$

Possible Answers:

$f'(x) = \cos(x)\cos(2x)+\sin(2x)\sin(x)$

$f'(x) = \cos(x)\cos(2x)-\sin(2x)\sin(x)$

$f'(x) = \cos(x)\cos(2x)-2\sin(2x)\sin(x)$

$f'(x) = \frac{\cos(x)\cos(2x)-2\sin(2x)\sin(x)}{\cos^2(2x)}$

Correct answer:

$f'(x) = \cos(x)\cos(2x)-2\sin(2x)\sin(x)$

Explanation:

To find the derivative of this function we must use the Product Rule and the Chain Rule. If $h(x) = \sin(x)$ and $g(x) = \cos(2x)$ then

$h'(x) = \cos(x)$

$g'(x) = -\sin(2x)(2) = -2\sin(2x)$

Applying these derivatives to the Product Rule gives us

$f'(x) = \cos(x)\cos(2x)-2\sin(2x)\sin(x)$

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Example Question #271 : Differential Functions

Differentiate:

$f(x) = \left(\frac{1}{x}\right)\cos^2(x)$

Possible Answers:

$f'(x) = \frac{-1}{x^2}\cos(x)-\frac{2}{x}\cos(x)\sin(x)$

$f'(x) = \frac{-1}{x^2}\cos(x)+\frac{2}{x}\cos(x)\sin(x)$

$f'(x) = \frac{-1}{x^2}\cos^2(x)-\frac{2}{x}\cos(x)\sin(x)$

$f'(x) = \frac{-\cos^2(x)-\cos(x)\sin(x)}{x^2}$

Correct answer:

$f'(x) = \frac{-1}{x^2}\cos^2(x)-\frac{2}{x}\cos(x)\sin(x)$

Explanation:

To find the derivative of this function we must use the Product Rule and the Chain Rule. First we set

$h(x) = \frac{1}{x}$

and

$g(x) = \cos^2(x)$

Now differentiating both of these functions gives

$h'(x) = \frac{-1}{x^2}$

$g'(x) = 2\cos(x)(-\sin(x)) = -2\cos(x)\sin(x)$

Applying this to the Product Rule gives us,

$f'(x) = \frac{-1}{x^2}\cos^2(x)-\frac{2}{x}\cos(x)\sin(x)$

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Example Question #272 : Differential Functions

Find the derivative of

$f(x) = \sin(x)\sin\left(\frac{1}{x}\right)$

Possible Answers:

$f'(x) = \cos(x)\sin\left(\frac{1}{x}\right)-\frac{\cos(\frac{1}{x})\sin(x)}{x^2}$

None of these answers are correct.

$f'(x) = \cos(x)\sin\left(\frac{1}{x}\right)-\frac{\cos(x)\sin(x)}{x^2}$

$f'(x) = \cos(x)\sin(x)-\frac{\cos(x)\sin(x)}{x^2}$

Correct answer:

$f'(x) = \cos(x)\sin\left(\frac{1}{x}\right)-\frac{\cos(\frac{1}{x})\sin(x)}{x^2}$

Explanation:

To find the derivative of this function we must use the Product Rule and the Chain Rule. If we have

$h(x) = \sin(x)$ and $g(x) = \sin\left(\frac{1}{x}\right)$ then

$h'(x) = \cos(x)$ and $g'(x) = \cos\left(\frac{1}{x}\right)\left(\frac{-1}{x^2}\right)$

Applying this to the product rule, we find

$f'(x) = \cos(x)\sin\left(\frac{1}{x}\right)+\cos\left(\frac{1}{x}\right)\sin(x)\left(\frac{-1}{x^2}\right) \rightarrow$

$f'(x) = \cos(x)\sin\left(\frac{1}{x}\right)-\frac{\cos(\frac{1}{x})\sin(x)}{x^2}$

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Example Question #271 : Functions

Find the derivative of the function

$f(x) = \frac{x^4\cos(3x)}{\sin(x)}$

Possible Answers:

$f'(x) = \frac{4x^3\cos(3x)\sin(x)-x^4\sin(3x)\sin(x)-x^4\cos(3x)\cos(x)}{\sin^2x}$

$f'(x) = \frac{4x^3\cos(3x)\sin(x)-3x^4\sin(3x)\sin(x)-x^4\cos(3x)\cos(x)}{\sin^2x}$

None of these answers are correct.

$f'(x) = \frac{4x^3\cos(3x)\sin(x)-3x^4\sin^2(3x)-x^4\cos^2(3x)}{\sin^2x}$

Correct answer:

$f'(x) = \frac{4x^3\cos(3x)\sin(x)-3x^4\sin(3x)\sin(x)-x^4\cos(3x)\cos(x)}{\sin^2x}$

Explanation:

To find the derivative of this function, we must use the Product Rule, Quotient Rule, and the Chain Rule. To do this, we first find the derivative of each part.

$h(x) = x^4 \Rightarrow h'(x) = 4x^3$

$g(x) = \cos(3x) \Rightarrow g'(x) = -3\sin(3x)$

$j(x) = \sin(x) \Rightarrow j'(x) = \cos(x)$

Using the derivatives of each part we can find the derivative of the numerator using the Product Rule

$h(x)g(x) = x^4\cos(3x) \Rightarrow \\ h'(x)g(x) + h(x)g'(x) = 4x^3\cos(3x)+(-3x^4)\sin(3x)$

Finally, putting this into the Quotient Rule gives

$f'(x) = \frac{[4x^3\cos(3x)+(-3x^4)\sin(3x)]\sin(x)-\cos(x)x^4\cos(3x)}{\sin^2x}$

$f'(x) = \frac{4x^3\cos(3x)\sin(x)-3x^4\sin(3x)\sin(x)-x^4\cos(3x)\cos(x)}{\sin^2x}$

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Example Question #274 : Differential Functions

Differentiate the function

$f(x) = \frac{\sin(x)}{x^2}$

Possible Answers:

$f'(x) = \frac{x^2\cos(x)+2x\sin(x)}{x^2}$

$f'(x) = \frac{x^2\cos(x)-2x\cos(x)}{x^4}$

$f'(x) = \frac{x^2\cos(x)-2x\sin(x)}{x^4}$

None of these answers are correct.

Correct answer:

$f'(x) = \frac{x^2\cos(x)-2x\sin(x)}{x^4}$

Explanation:

To differentiate this function we must use the Quotient Rule

$f'(x) = \frac{h'(x)g(x)-h(x)g'(x)}{(g(x))^2}$

Using $h(x) = \sin(x)$ and g(x) = x^2 .

The derivative of the function is then

$f'(x) = \frac{x^2\cos(x)-2x\sin(x)}{x^4}$

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Example Question #82 : How To Find Differential Functions

Differentiate the following function

$f(x) = \sin^2(\cos(2x))$

Possible Answers:

None of these answers are correct.

$f'(x) = 2\sin(\cos(2x))\cos(\cos(2x))\sin(2x)$

$f'(x) = -4\cos(\cos(2x))\cos(\sin(2x))\sin(2x)$

$f'(x) = -4\sin(\cos(2x))\cos(\cos(2x))\sin(2x)$

Correct answer:

$f'(x) = -4\sin(\cos(2x))\cos(\cos(2x))\sin(2x)$

Explanation:

To differentiate this function we must use the Chain Rule. where

$f(x) = h(g(x)) \Rightarrow f'(x) = h'(g(x))\cdot g'(x)$

Therefore the derivative of the function is

$f(x) = 2\sin(\cos(2x))\cos(\cos(2x))(-\sin(2x))(2) \rightarrow$

$f'(x) = -4\sin(\cos(2x))\cos(\cos(2x))\sin(2x)$

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Example Question #82 : Other Differential Functions

Find the derivative of the function

$f(x) = x^2\cos(x)$

Possible Answers:

$f'(x) = 2x\cos(x) + x^2\sin(x)$

None of these answers are correct.

$f'(x) = 2x\sin(x) + x^2\sin(x)$

$f'(x) = 2x\cos(x) - x^2\sin(x)$

Correct answer:

$f'(x) = 2x\cos(x) - x^2\sin(x)$

Explanation:

To find the derivative of the function, we must use the Product Rule,

f'(x) = h'(x)g(x)+h(x)g'(x)

Using h(x) = x^2 and $g(x) = \cos(x)$ .

Therefore

$f'(x) = 2x\cos(x) + x^2(-\sin(x)) = 2x\cos(x) -x^2\sin(x)$

$f'(x) = 2x\cos(x) - x^2\sin(x)$

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Example Question #91 : Other Differential Functions

Differentiate the function

$f(x) = \sin(\cos(2x))$

Possible Answers:

None of these answers are correct.

$f(x) = \cos(\cos(2x))\sin(2x)$

$f(x) = 2\cos(\cos(2x))\sin(2x)$

$f(x) = -2\cos(\cos(2x))\sin(2x)$

Correct answer:

$f(x) = -2\cos(\cos(2x))\sin(2x)$

Explanation:

To differentiate the function properly, we must use the Chain Rule which is,

$f(x) = h(g(x)) \rightarrow f'(x) = h'(g(x))g'(x)$

Therefore the derivative of the function is,

$f'(x) = \cos(\cos(2x))(-\sin(2x))(2) = -2\cos(\cos(2x))\sin(2x)$

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Example Question #92 : Other Differential Functions

Differentiate

$f(x) = \sin^2(x^2)$

Possible Answers:

$f'(x) = -2\sin(x^2)\cos(x^2)(2x)$

$f'(x) = -4\sin(x^2)\cos(x^2)(2x)$

$f'(x) = 4\sin(x^2)\cos(x^2)(2x)$

$f'(x) = 4x\sin(x^2)\cos(x^2)$

Correct answer:

$f'(x) = 4x\sin(x^2)\cos(x^2)$

Explanation:

To differentiate this equation we use the Chain Rule.

$f(x) = h(g(x)) \rightarrow f'(x) = h'(g(x))\cdot g'(x)$

Using this throughout the equation gives us,

$f'(x) = 2\sin(x^2)\cos(x^2)(2x)$

Report an Error

Example Question #271 : Differential Functions

Find the derivative of

$f(x) = 4\sin(x+3x^2)$

Possible Answers:

$\small f'(x) = 4\cos(x+3x^2)$

None of these answers are correct.

$\small f'(x) = -4\cos(x+3x^2)(1+6x)$

$\small f'(x) = 4\cos(x+3x^2)(1+6x)$

Correct answer:

$\small f'(x) = 4\cos(x+3x^2)(1+6x)$

Explanation:

To find the derivative of the function we must use the Chain Rule, which is

$f(x) = h(g(x)) \rightarrow f'(x) = h'(g(x))\cdot g'(x)$

Applying this to the function we get,

$f'(x) = 4\cos(x+3x^2)(1+6x)$

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Calculus 1 : Differential Functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #81 : Other Differential Functions

Example Question #271 : Differential Functions

Example Question #272 : Differential Functions

Example Question #271 : Functions

Example Question #274 : Differential Functions

Example Question #82 : How To Find Differential Functions

Example Question #82 : Other Differential Functions

Example Question #91 : Other Differential Functions

Example Question #92 : Other Differential Functions

Example Question #271 : Differential Functions

All Calculus 1 Resources

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