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Calculus 1 : Differential Functions

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Example Questions

Example Question #74 : How To Find Differential Functions

Determine all values which result in critical points for the following function:

$f(x)=e^{x^3-2x^2-7x}$

Possible Answers:

$x=\frac{7}{3}, x=-1$

$x=-\frac{7}{3},x=1$

$x=-2, x=\frac{1}{5}$

$x=2,x=-\frac{1}{5}$

x=3, x=-1

Correct answer:

$x=\frac{7}{3}, x=-1$

Explanation:

The critical points of the function are those at which its slope is 0, so to find the critical points we must first take the derivative of the function:

$f'(x)=(3x^2-4x-7)e^{x^3-2x^2-7x}$

The points where the slope of the function is 0 are those where the derivative is 0, so we then factor the first term in the equation above to make it easier to see at what values of x the slope of the function is 0:

$f'(x)=(3x-7)(x+1)e^{x^3-2x^2-7x}$

We can see from the equation for our derivative that only the first and second terms can make the slope equal zero, so we set each of these terms equal to zero and solve for x to find where our critical points would be:

$3x-7=0\rightarrow x=\frac{7}{3}$

$x+1=0\rightarrow x=-1$

Remember critical points are the coordinate pair (x, f(x)) we would need to plug in our x values into the original function to find the cooresponding y values. However, this question only asks to find the x values of the critical points.

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Example Question #75 : How To Find Differential Functions

Find the derivative of the following function:

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

Possible Answers:

$\ln(x)+4x$

$\ln(x)+8x+8$

$\frac{1}{x}+8x+8$

$\frac{1}{x}+4x+8$

Correct answer:

$\frac{1}{x}+8x+8$

Explanation:

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

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

From here we use the power rule which states when,

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

We also need to remember,

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

Therefore our derivative becomes,

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

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Example Question #76 : How To Find Differential Functions

Find the derivative of the following function:

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

Possible Answers:

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

None of these answers are correct.

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

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

Correct answer:

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

Explanation:

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

$\frac{dy}{dx}[x^2] = 2x$

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

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

Knowing this we can differentiate the function

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

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

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Example Question #77 : How To Find Differential Functions

Find the derivative of the following function

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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

g(x) = cos(x)

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

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

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

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

Combining everything together gives us

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

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Example Question #73 : How To Find Differential Functions

Find the derivative of the following function

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

The Quotient Rule is

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

The Chain Rule is

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

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

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

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

Taking the derivative of the denominator gives us

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

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

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

Simplifying this gives us

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

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Example Question #79 : How To Find Differential Functions

Find the derivative of the following function

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

Possible Answers:

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

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

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

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

Correct answer:

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

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

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

The equation for the Chain Rule is

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

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

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

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

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

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Example Question #261 : Differential Functions

Find the derivative of the following function

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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

Applying this to the function we are given gives us

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

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

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Example Question #261 : Differential Functions

Find the derivative of the function

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

Possible Answers:

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

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

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

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

Correct answer:

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

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

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

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

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

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Example Question #261 : Functions

Find the derivative of

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

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

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Example Question #81 : How To Find Differential Functions

Find the derivative of the following function

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

Possible Answers:

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

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

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

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

Correct answer:

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

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

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

Using this with the Quotient Rule, we find

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

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

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Calculus 1 : Differential Functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #74 : How To Find Differential Functions

Example Question #75 : How To Find Differential Functions

Example Question #76 : How To Find Differential Functions

Example Question #77 : How To Find Differential Functions

Example Question #73 : How To Find Differential Functions

Example Question #79 : How To Find Differential Functions

Example Question #261 : Differential Functions

Example Question #261 : Differential Functions

Example Question #261 : Functions

Example Question #81 : How To Find Differential Functions

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