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Calculus 1 : How to find differential functions

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Example Questions

Example Question #811 : Differential Functions

Find $\frac{dy}{dx}$ for the follow function:

y = x^y

Possible Answers:

$\frac{dy}{dx}= \frac{y}{(\frac{1}{y}-yln(x))}$

$\frac{dy}{dx}= \frac{xy}{(\frac{1}{y}-ln(x))}$

$\frac{dy}{dx}= \frac{y}{x(1-ln(x))}$

$\frac{dy}{dx}= \frac{y}{x(y-ln(x))}$

$\frac{dy}{dx}= \frac{y}{x(\frac{1}{y}-ln(x))}$

Correct answer:

$\frac{dy}{dx}= \frac{y}{x(\frac{1}{y}-ln(x))}$

Explanation:

To make solving easier, take the natural log of both sides of the equations:

$ln(y) = ln(x^{y})$

Use the natural log properties of ln(a^b) = bln(a) .

ln(y) = yln(x)

Now take the derivative of both sides of the equation, note that the derivative of is $\frac{dy}{dx}$ in this case. Must also use the power rule for yln(x) . The general equation is $\frac{d(ab)}{dx} = \frac{da}{dx}b+a\frac{db}{dx}$

Applying the power rule:
$\frac{dy}{dx}\frac{1}{y} = \frac{dy}{dx}ln(x)+\frac{y}{x}$

Now simplify for $\frac{dy}{dx}$ :

$\frac{dy}{dx}\frac{1}{y} -\frac{dy}{dx}ln(x)= \frac{y}{x}-->\frac{dy}{dx}(\frac{1}{y}-ln(x))= \frac{y}{x}$

$\frac{dy}{dx}= \frac{y}{x(\frac{1}{y}-ln(x))}$

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Example Question #812 : Differential Functions

Find $\frac{dy}{dx}$ for the function below:

y = 10^x

Hint: Use implicit differentiation.

Possible Answers:

$\frac{dy}{dx} = 10^xln(x)$

$\frac{dy}{dx} = 10ln(x)$

$\frac{dy}{dx} = 10xln(10x)$

$\frac{dy}{dx} = 10^xln(10)$

$\frac{dy}{dx} = 10^xln(10x)$

Correct answer:

$\frac{dy}{dx} = 10^xln(10)$

Explanation:

To make solving easier, take the natural log of both sides of the equations:

ln(y) = ln(10^x)

Use the natural log properties of ln(a^b) = bln(a)

ln(y) = xln(10)

Now take the derivative of both sides of the equation, note that the derivative of is $\frac{dy}{dx}$ in this case.

$\frac{dy}{dx}\frac{1}{y}= ln(10) -->\frac{dy}{dx} = yln(10)$

Plug the definition of back into:

$\frac{dy}{dx} = 10^xln(10)$

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Example Question #1842 : Calculus

Find $\frac{dy}{dx}$ for the function below:

$y = \frac{(x+1)^5}{(x+3)^2}$

Hint: Use implicit differentiation.

Possible Answers:

$\frac{dy}{dx}=y\frac{3x+13}{(x+1)(x+3)}$

$\frac{dy}{dx}=y\frac{(x+2)}{(x+1)(x+3)}$

$\frac{dy}{dx}=y\frac{(x+2)}{(x+5)(x+3)}$

$\frac{dy}{dx}=y\frac{6(x+4)}{(x+6)(x+3)}$

$\frac{dy}{dx}=y\frac{(x+2)}{(x+4)(x+3)}$

Correct answer:

$\frac{dy}{dx}=y\frac{3x+13}{(x+1)(x+3)}$

Explanation:

To make solving easier, take the natural log of both sides of the equations:

$ln(y) = ln[\frac{(x+1)^5}{(x+3)^2}]$

Use the natural log properties of ln(a^b) = bln(a) and $ln(\frac{a}{b}) = ln(a)-ln(b)$

$ln(y) = ln[\frac{(x+1)^5}{(x+3)^2}] = ln((x+1)^5)-ln((x+3)^2)$

ln(y)= 5ln(x+1)-2ln(x+3)

Now take the derivative of both sides of the equation, note that the derivative of is $\frac{dy}{dx}$ in this case. The derivative of ln(a) is $\frac{1}{a}\frac{da}{dx}$

$\frac{dy}{dx} \frac{1}{y}= \frac{5}{x+1}-\frac{2}{x+3}$

Simplify and combine both the terms under a common denominator.

$\frac{dy}{dx} = y(\frac{5}{x+1}-\frac{2}{x+3}) = y\frac{5(x+3)-2(x+1)}{(x+1)(x+3)}$

$\frac{dy}{dx}= y\frac{5(x+3)-2(x+1)}{(x+1)(x+3)} =y\frac{3x+13}{(x+1)(x+3)}$

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Example Question #821 : Differential Functions

Find $\frac{dy}{dx}$ for the function below:

$y = x^{cos(x)}$

Hint: Use implicit differentiation.

Possible Answers:

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

$\frac{dy}{dx} = y[tan(x)ln(x)+\frac{cos(x)}{x}]$

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

$\frac{dy}{dx} = y[ln(x)+\frac{cos(x)}{x}]$

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

Correct answer:

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

Explanation:

To make solving easier, take the natural log of both sides of the equations:

$ln(y) = ln(x^{cos(x)})$

Use the natural log properties of ln(a^b) = bln(a).

ln(y) = cos(x)ln(x)

Now take the derivative of both sides of the equation, note that the derivative of is $\frac{dy}{dx}$ in this case. Must also use the power rule for the cos(x)ln(x) term. The general equation is $\frac{d(ab)}{dx} = \frac{da}{dx}b+a\frac{db}{dx}$

$\frac{dy}{dx} \frac{1}{y}= -sin(x)ln(x)+cos(x)\frac{1}{x}$

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

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Example Question #631 : How To Find Differential Functions

Find $\frac{dy}{dx}$ for the function below:

y = ln(xy)

Hint: Use implicit differentiation.

Possible Answers:

$\frac{dy}{dx} =\frac{x}{y-1}$

$\frac{dy}{dx} =\frac{y}{x(y-1)}$

$\frac{dy}{dx} =\frac{x-1}{y-1}$

$\frac{dy}{dx} =\frac{y}{x}$

$\frac{dy}{dx} =\frac{y}{xy-1}$

Correct answer:

$\frac{dy}{dx} =\frac{y}{x(y-1)}$

Explanation:

To take the derivative of any ln(a) term:

Step 1: Take whatever is inside the and divide it by 1.

$\frac{1}{xy}$

Step 2: Take the derivative of the inside and multiply it by step 1: Don't for get to use the product rule and remember that the derivative of is $\frac{dy}{dx}$ .

Product Rule:

$\frac{d}{dx}(xy) = x\frac{dy}{dx} +(1)y$

Multiply the step 1 product by the step 2 product.

$\frac{dy}{dx} = \frac{1}{xy}(x\frac{dy}{dx} +(1)y)$

Now isolate just for dy/dx

$\frac{dy}{dx} = \frac{1}{xy}(x\frac{dy}{dx} +(1)y)=\frac{dy}{dx} \frac{1}{y}+\frac{1}{x}$

Place the $\frac{dy}{dx}$ terms on the same sides.

$\frac{dy}{dx }- \frac{dy}{dx} \frac{1}{y}= \frac{1}{x}$

$\frac{dy}{dx }(1-\frac{1}{y})= \frac{1}{x}$

Divide both sides by the term attached to the $\frac{dy}{dx}$ and simplify.

$\frac{dy}{dx }= \frac{1}{x(1-\frac{1}{y})} = \frac{1}{x(\frac{y-1}{y})}= \frac{y}{x(y-1)}$

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Example Question #632 : How To Find Differential Functions

Find $\frac{dy}{dx}$ for the function below:

y^3+y^2+y+x = 0

Hint: Use implicit differentiation.

Possible Answers:

$\frac{dy}{dx}=\frac{-1}{3y^2+2y+1}$

$\frac{dy}{dx}=\frac{4}{y+1}$

$\frac{dy}{dx}=\frac{1}{y^2+y+1}$

$\frac{dy}{dx}=\frac{3y}{y+1}$

$\frac{dy}{dx}=\frac{y+2}{y+1}$

Correct answer:

$\frac{dy}{dx}=\frac{-1}{3y^2+2y+1}$

Explanation:

Take the derivative of each term and remember that the derivative of is $\frac{dy}{dx}$ .

$\frac{d}{dx}(y^3+y^2+y+x = 0) -->3y^2\frac{dy}{dx}+2y\frac{dy}{dx}+\frac{dy}{dx}+1=0$

Now isolate for $\frac{dy}{dx}$

$\frac{dy}{dx}(3y^2+2y+1)=-1$

$\frac{dy}{dx}=\frac{-1}{3y^2+2y+1}$

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Example Question #633 : How To Find Differential Functions

Find $\frac{dy}{dx}$ for the follow function:

y = ln(x+y)

Possible Answers:

$\frac{dy}{dx}=\frac{1}{(x-y-1)}$

$\frac{dy}{dx}=\frac{1}{(x+y+1)}$

$\frac{dy}{dx}=\frac{xy}{(x-y-1)}$

$\frac{dy}{dx}=\frac{x+y}{(x+y+1)}$

$\frac{dy}{dx}=\frac{1}{(x+y-1)}$

Correct answer:

$\frac{dy}{dx}=\frac{1}{(x+y-1)}$

Explanation:

To take the derivative of any ln(x) term:

Step 1: Take whatever is inside the and divide it by 1.

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

Step 2: Take the derivative of the inside and multiply it by step 1:

$\frac{d}{dx}(x+y)= 1+\frac{dy}{dx}$

Put the $\frac{dy}{dx}$ terms together to simplify:

$\frac{dy}{dx}(x+y)-\frac{dy}{dx}=1$

Isolate for $\frac{dy}{dx}$ :

$\frac{dy}{dx}(x+y-1)=1$

$\frac{dy}{dx}=\frac{1}{(x+y-1)}$

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Example Question #634 : How To Find Differential Functions

Find the first derivative of the following function:

f(x)=(2x+4)(x^2-5)

Possible Answers:

f'(x)=(2x+4)(2x)+(2)(x^2-5)

f'(x)=(2x+4)(2x)-(2)(x^2-5)

f'(x)=0

f'(x)=4x

Correct answer:

f'(x)=(2x+4)(2x)+(2)(x^2-5)

Explanation:

To solve, simply use the product rule and power rule for derivatives.

Product rule: (g(x)h(x))'=g(x)h'(x)+g'(x)h(x)

Power rule: $(x^n)'=nx^{n-1}$

where

g(x)=2x+4

h(x)=x^2-5

Thus,

g'(x)=2

h'(x)=2x

Therefore,

f'(x)=(2x+4)(2x)+(2)(x^2-5)

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Example Question #635 : How To Find Differential Functions

Determine the slope of the line that is normal to the function $f(x)=\frac{x^3+15x+4}{3}$ at the point x=-2

Possible Answers:

$-\frac{1}{9}$

$\frac{1}{9}$

Correct answer:

$-\frac{1}{9}$

Explanation:

A line that is normal, that is to say perpendicular to a function at any given point will be normal to this slope of the line tangent to the function at that point.

The slope of the tangent can be found by taking the derivative of the function and evaluating the value of the derivative at a point of interest.

Taking the derivative of the function $f(x)=\frac{x^3+15x+4}{3}$ at the point x=-2

The slope of the tangent is

$f'(x)=\frac{3x^2+15}{3}=x^2+5$

f'(-2)=(-2)^2+5=9

Since the slope of the normal line is perpendicular, it is the negative reciprocal of this value

$m=\frac{-1}{9}$

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Example Question #636 : How To Find Differential Functions

Determine the slope of the line that is normal to the function f(x)=3sin^2(x) at the point $x=\frac{\pi}{4}$

Possible Answers:

$-\frac{1}{3}$

$\frac{1}{3}$

$-\frac{1}{6}$

Correct answer:

$-\frac{1}{3}$

Explanation:

A line that is normal, that is to say perpendicular to a function at any given point will be normal to this slope of the line tangent to the function at that point.

The slope of the tangent can be found by taking the derivative of the function and evaluating the value of the derivative at a point of interest.

We'll need to make use of the following derivative rule(s):

Trigonometric derivative:

d[sin(u)]=cos(u)du

Note that u may represent large functions, and not just individual variables!

Taking the derivative of the function f(x)=3sin^2(x) at the point $x=\frac{\pi}{4}$

The slope of the tangent is

f'(x)=6sin(x)cos(x)

$f'(\frac{\pi}{4})=6sin(\frac{\pi}{4})cos(\frac{\pi}{4})=3$

Since the slope of the normal line is perpendicular, it is the negative reciprocal of this value

$m=-\frac{1}{3}$

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Calculus 1 : How to find differential functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #811 : Differential Functions

Example Question #812 : Differential Functions

Example Question #1842 : Calculus

Example Question #821 : Differential Functions

Example Question #631 : How To Find Differential Functions

Example Question #632 : How To Find Differential Functions

Example Question #633 : How To Find Differential Functions

Example Question #634 : How To Find Differential Functions

Example Question #635 : How To Find Differential Functions

Example Question #636 : How To Find Differential Functions

All Calculus 1 Resources

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