Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

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Example Questions

Example Question #72 : Other Differential Functions

Find the derivative of the following function:

\(\displaystyle f(x)=\ln(x)+4x(x+2)\)

Possible Answers:

\(\displaystyle \frac{1}{x}+8x+8\)

\(\displaystyle \ln(x)+8x+8\)

\(\displaystyle \frac{1}{x}+4x+8\)

\(\displaystyle \ln(x)+4x\)

Correct answer:

\(\displaystyle \frac{1}{x}+8x+8\)

Explanation:

To take the derivative of this function lets first distribute the 4x through the binomial.

\(\displaystyle f(x)=\ln(x)+4x^2+8x\)

From here we use the power rule which states when,

\(\displaystyle f(x)=x^n\rightarrow f'(x)=nx^{n-1}\).

We also need to remember,

\(\displaystyle \frac{d}{dx}\ln(x)=\frac{1}{x}\).

Therefore our derivative becomes,

\(\displaystyle f'(x)=\frac{1}{x}+8x+8\).

Example Question #73 : Other Differential Functions

Find the derivative of the following function:

\(\displaystyle f(x) = \sin^2(x) - \sin(\cos(x))\)

Possible Answers:

\(\displaystyle f(x) = 2\sin(x)\cos(x) + \cos(\cos(x))\sin(x)\)

None of these answers are correct.

\(\displaystyle f(x) = 2\sin(x) + \cos(x)\sin(x)\)

\(\displaystyle f(x) = \cos^2(x) + \cos(x)\sin(x)\)

Correct answer:

\(\displaystyle f(x) = 2\sin(x)\cos(x) + \cos(\cos(x))\sin(x)\)

Explanation:

To find the derivative of the function, we must use the Chain Rule. Since \(\displaystyle \sin(x)\) and \(\displaystyle \cos(x)\) are both functions, we can 

\(\displaystyle \frac{dy}{dx}[x^2] = 2x\)

\(\displaystyle \frac{dy}{dx}[\sin(x)] = \cos(x)\)

\(\displaystyle \frac{dy}{dx}[\cos(x)] =- \sin(x)\)

Knowing this we can differentiate the function

\(\displaystyle \frac{dy}{dx} = 2\sin(x)\cdot \frac{dy}{dx}[\sin(x)] -\cos(\cos(x))\cdot\frac{dy}{dx}[cos(x)]\)

\(\displaystyle \frac{dy}{dx} = 2\sin(x)\cos(x) +\cos(\cos(x))\sin(x)\)

Example Question #74 : Other Differential Functions

Find the derivative of the following function

\(\displaystyle f(x) = \sin(x^2+5x)\cos(x)\)

Possible Answers:

\(\displaystyle {f}'(x) = \cos(x^2+5x)(2x+5)\cos(x)-\sin(x)\sin(x^2+5x)\)

\(\displaystyle {f}'(x) = \cos(x^2+5x)\cos(x)-\sin(x)\sin(x^2+5x)\)

\(\displaystyle {f}'(x) = (2x-5)\cos(x^2+5x)\cos(x)-\sin(x)\sin(x^2+5x)\)

\(\displaystyle {f}'(x) = (2x+5)\sin(x^2)\sin(5x)\cos(x)-\sin(x)\sin(x^2+5x)\)

Correct answer:

\(\displaystyle {f}'(x) = \cos(x^2+5x)(2x+5)\cos(x)-\sin(x)\sin(x^2+5x)\)

Explanation:

We must use the Product Rule and the Chain Rule to find the derivative of this function. If we let

\(\displaystyle h(x) = \sin(x^2+5x)\)

\(\displaystyle g(x) = cos(x)\)

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

Now we can find the derivative of the two parts using the Chain Rule

\(\displaystyle h'(x) = \cos(x^2+5x)(2x+5)\)

\(\displaystyle g'(x) = -\sin(x)\)

Combining everything together gives us

\(\displaystyle f'(x) = \cos(x^2+5x)(2x+5)(\cos(x))-\sin(x))(\sin(x^2+5x))\)

 

Example Question #75 : Other Differential Functions

Find the derivative of the following function

\(\displaystyle f(x) = \frac{\sin(x^2)}{2x^3}\)

Possible Answers:

\(\displaystyle f'(x) = \frac{\cos(x^2)(4x^4)+6x^2\sin(x^2)}{4x^6}\)

\(\displaystyle f'(x) = \frac{\cos(x^2)(2x^4)-6x^2\sin(x^2)}{4x^6}\)

\(\displaystyle f'(x) = \frac{\cos(x^2)(2x^4)-6x^2\sin(x^2)}{2x^6}\)

\(\displaystyle f'(x) = \frac{\cos(x^2)(4x^4)-6x^2\sin(x^2)}{4x^6}\)

Correct answer:

\(\displaystyle f'(x) = \frac{\cos(x^2)(4x^4)-6x^2\sin(x^2)}{4x^6}\)

Explanation:

To find the derivative of the function we must use the Chain Rule and the Quotient Rule.

The Quotient Rule is 

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

The Chain Rule is 

\(\displaystyle f'(x) = h'(g(x))\cdot g(x)\)

If \(\displaystyle h(x) = \sin(x^2)\) and \(\displaystyle g(x) = 2x^3\)

Using the Chain Rule on the numerator of the function gives us,

\(\displaystyle h'(x) = \cos(x^2)*(2x)\)

Taking the derivative of the denominator gives us

\(\displaystyle g'(x) = (3\cdot 2)x^{3-1} = 6x^2\)

Now we can combine all of the pieces into the Quotient Rule to find the derivative of the entire function

\(\displaystyle f'(x) = \frac{\cos(x^2)(2x)(2x^3)-6x^2\sin(x^2)}{(2x^3)^2}\)

Simplifying this gives us

\(\displaystyle f'(x) = \frac{\cos(x^2)(4x^4)-6x^2\sin(x^2)}{4x^6}\)

Example Question #76 : Other Differential Functions

Find the derivative of the following function

\(\displaystyle f(x) = x^2\sin\left(\frac{1}{x}\right)\)

Possible Answers:

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

\(\displaystyle f'(x) = 2x\sin\left(\frac{1}{x}\right) + x^2\sin\left(\frac{1}{x}\right)\)

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

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

Correct answer:

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

Explanation:

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

The equation for the Product Rule is

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

The equation for the Chain Rule is 

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

Applying the Chain Rule to \(\displaystyle \sin\left(\frac{1}{x}\right)\) gives

\(\displaystyle \cos\left(\frac{1}{x}\right)\left(\frac{-1}{x^2}\right)\)

Using the Product Rule, we can now find the derivative of the entire function. If \(\displaystyle h(x) = x^2\) and \(\displaystyle g(x) = \sin\left(\frac{1}{x}\right)\) then the derivative of the function will be,

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

Example Question #77 : Other Differential Functions

Find the derivative of the following function

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

Possible Answers:

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

\(\displaystyle f'(x) = -8x\cos(\sin(x^2))\sin(\sin(x^2))\cos(x^2)\)

\(\displaystyle f'(x) = -8x\cos(\sin(x^2))\sin(\sin(x^2))\cos(x^2)(2x)\)

\(\displaystyle f'(x) = -8x\cos^2(\sin^2(\sin(x^2))\)

Correct answer:

\(\displaystyle f'(x) = -8x\cos(\sin(x^2))\sin(\sin(x^2))\cos(x^2)\)

Explanation:

To find the derivative of this equation, we must use the Chain Rule.

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

Applying this to the function we are given gives us

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

\(\displaystyle f'(x) = -8x\cos(\sin(x^2))\sin(\sin(x^2))\cos(x^2)\)

Example Question #261 : Functions

Find the derivative of the function

\(\displaystyle f(x) = \frac{\cos(\sin(x))}{\sin(x)}\)

Possible Answers:

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

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

\(\displaystyle f'(x) = \frac{-\sin(\sin(x))\cos(x)\sin(x)-\cos(x)\cos(\sin(x))}{\sin(x)}\)

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

Correct answer:

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

Explanation:

To find the derivative of this function we must use the Chain Rule and the Quotient Rule. Applying the Chain Rule to the numerator gives

\(\displaystyle h'(x) = -\sin(\sin(x))\cos(x)\)

Now using the Quotient Rule for the function, we find the derivative to be

\(\displaystyle f'(x) = \frac{[-\sin(\sin(x))\cos(x)]\sin(x)-\cos(x)\cos(\sin(x))}{\sin^2(x)}\)

Example Question #82 : Other Differential Functions

Find the derivative of 

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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

Applying this to the function we are given, with \(\displaystyle h(x) = x^2\) and \(\displaystyle g(x) = \sin(x)\) gives us 

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

Example Question #83 : Other Differential Functions

Find the derivative of the following function

\(\displaystyle f(x) = \frac{x\sin(x)}{\cos(x)}\)

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

To find the derivative of this function we must use the Product Rule and the Quotient Rule. Appling the Product Rule to the numerator of the function gives us

\(\displaystyle h'(x) = \sin(x)+x\cos(x)\)

Using this with the Quotient Rule, we find 

\(\displaystyle f'(x) = \frac{[\sin(x)+x\cos(x)]\cos(x)-x\sin(x)(-\sin(x))}{\cos^2(x)} \rightarrow\)

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

Example Question #84 : Other Differential Functions

Find the derivative of 

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

Applying these derivatives to the Product Rule gives us

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

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