Log and Exponential Tangents and Normals

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Question 1 of 4

Find the equation of the tangent to the curve

$y=ln(5x+5)$ at the point

$x=3$ .

1. $4y-4ln20-x+3=0$
2. $4y-4ln20=0$
3. $y-4ln20-x=0$
4. $4y+4ln20-x+3=0$

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Derivative of $y=ln f(x)$

$y’=\frac{\color{#9a00c7}{f'(x)}}{\color{#D800AD}{f(x)}}$

Point Gradient Formula

$y-$ $y_1$

$=$ $m$

$(x-$ $x_1$

$)$

First, substitute $x=3$ to the function to get the value of $y$ .

$y$	$=$	$ln(5$ $x$ $+5)$
	$=$	$ln(5$ ( $3$ $)+5)$	Substitute known values
	$=$	$ln20$

To find the slope, find the derivative of the function then substitute the value of

$x$

$x$	$=$	$3$
$f(x)$	$=$	$5x+5$

$m$	$=$	$y’$

	$=$	$\frac{\color{#9a00c7}{f'(x)}}{\color{#D800AD}{f(x)}}$	$y’=(f'(x))/(f(x))$

	$=$	$\frac{\color{#9a00c7}{f'(5x+5)}}{\color{#D800AD}{5x+5}}$	Substitute known values

	$=$	$\frac{5}{5(\color{#e65021}{x})+5}$	Differentiate $5x+5$

	$=$	$\frac{5}{5(\color{#e65021}{3})+5}$	Substitute known values

	$=$	$5/(20)$

	$=$	$1/4$	Simplify

Next, substitute the components to the point gradient formula.

$x_1$	$=$	$3$
$y_1$	$=$	$ln20$

$m$	$=$	$1/4$

$y-$ $y_1$	$=$	$m$ $(x-$ $x_1$ $)$

$y-$ $ln20$	$=$	$1/4$ $(x-$ $3$ $)$	Substitute known values

$(y-ln20)$ $times4$	$=$	$(1/4(x-3))$ $times4$	Multiply both sides by $4$

$4y-4ln20$	$=$	$x-3$
$4y-4ln20-x+3$	$=$	$0$	Move all terms to the left side

$4y-4ln20-x+3=0$

Question 2 of 4

Given the function

$y=e^(2x)$ , find the equation of the tangent line to the point

$(2,e^4)$

1. $y=2xe^2-3e^4$
2. $y=-3e^4$
3. $y=2xe^4$
4. $y=2xe^4-3e^4$

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Product Rule with Base “e”

$y’=\color{#9a00c7}{f'(x)}\cdot e^{\color{#D800AD}{f(x)}}$

Point Gradient Formula

$y-$ $y_1$

$=$ $m$

$(x-$ $x_1$

$)$

To find the slope, find the derivative of the function then substitute the value of

$x$

$=$

$2$

$m$	$=$	$y’$
	$=$	$\color{#9a00c7}{f'(x)}\cdot e^{\color{#D800AD}{f(x)}}$	$y’=f'(x) e^(f(x))$
	$=$	$\color{#9a00c7}{f'(2x)}\cdot e^{\color{#D800AD}{2x}}$	Substitute known values
	$=$	$2\cdot e^{2\color{#e65021}{x}}$	Differentiate $2x$
	$=$	$2e^{2(\color{#e65021}{2})}$	Substitute known values
	$=$	$2e^4$

Next, substitute the components to the point gradient formula.

$x_1$	$=$	$2$
$y_1$	$=$	$e^4$
$m$	$=$	$2e^4$

$y-$ $y_1$	$=$	$m$ $(x-$ $x_1$ $)$
$y-$ $e^4$	$=$	$2e^4$ $(x-$ $2$ $)$	Substitute known values
$y-e^4$	$=$	$2xe^4-4e^4$	Distribute
$y-e^4$ $+e^4$	$=$	$2xe^4-4e^4$ $+e^4$	Add $e^4$ to both sides
$y$	$=$	$2xe^4-3e^4$

$y=2xe^4-3e^4$

Question 3 of 4

3. Question
Find the slope ( $m$ ) and the angle of inclination ( $theta$ ) at:

$x=4$

Round your answer to two decimal places
- $m=$ (54.98)
  
  $theta=$ (89) $°$
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Slope of an Exponential Function

$y’=e^x$

To find the slope, find the derivative of the function then substitute the value of $x$

$m$ $=$ $y’$

$=$ $e^x$ $y’=e^x$

$=$ $e^4$ Substitute known values

$=$ $54.98$ Compute using calculator

Next, solve for the angle of inclination using $tantheta=$ $m$ , where $theta$ is the angle of inclination.

$tantheta$ $=$ $m$

$theta$ $=$ $tan^-1$ $(m)$ Reverse the function to get $theta$

$theta$ $=$ $tan^-1$ $(54.98)$ Substitute known values

$theta$ $=$ $89°$ Compute using calculator

$m=54.98$

$theta=89°$

Question 4 of 4

Find the equation of the normal line that intersects the curve

$y=ln sqrtx$ at the point

$y=-1$ .

1. $y=-2/(e^2)x-1$
2. $y=2/(e^2)x+2/(e^4)+1$
3. $y=-2/(e^4)x+2/(e^4)-1$
4. $y=-2/(e^2)x+2/(e^4)-1$

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Derivative of $y=ln f(x)$

$y’=\frac{\color{#9a00c7}{f'(x)}}{\color{#D800AD}{f(x)}}$

Point Gradient Formula

$y-$ $y_1$

$=$ $m$

$(x-$ $x_1$

$)$

First, substitute $y=-1$ to the function to get the value of $x$ .

$y$	$=$	$ln sqrtx$
$-1$	$=$	$ln sqrtx$	Substitute known values

$-1$	$=$	$\ln x^{\color{#CC0000}{\frac{1}{2}}}$	Change the surd into an exponent

$\color{#CC0000}{e}^{-1}$	$=$	$\color{#CC0000}{e}^{\ln x^{\frac{1}{2}}}$	Insert $e$ as base to both sides

$e^(-1)$	$=$	$x^{\frac{1}{2}}$	$e^(ln x)=x$

$(e^{-1})^\color{#CC0000}{2}$	$=$	$(x^{\frac{1}{2}})^\color{#CC0000}{2}$	Raise both sides by $2$

$e^(-2)$	$=$	$x$

$x$	$=$	$1/(e^2)$	Reciprocate $e^(-2)$

To find the tangent, find the derivative of the function then substitute the value of

$x$

$x$	$=$	$1/(e^2)$

$f(x)$	$=$	$x^(1/2)$

$y’$	$=$	$\frac{\color{#9a00c7}{f'(x)}}{\color{#D800AD}{f(x)}}$

	$=$	$\frac{\color{#9a00c7}{f'(x^{\frac{1}{2}})}}{\color{#D800AD}{x^{\frac{1}{2}}}}$	Substitute known values

	$=$	$\frac{\frac{1}{2}x^{-\frac{1}{2}}}{x^{\frac{1}{2}}}$	Differentiate $x^(1/2)$

	$=$	$1/(x^(1/2)) times 1/(2x^(1/2))$	Simplify

	$=$	$\frac{1}{2\color{#e65021}{x}}$

	$=$	$\frac{1}{2\color{#e65021}{\frac{1}{e^2}}}$	Substitute known values

	$=$	$1/(2/(e^2))$

	$=$	$(e^2)/2$	Reciprocate $2/(e^2)$

To find the normal line, get the negative reciprocal of the tangent, which is

$(e^2)/2$

	$(e^2)/2$

$=$	$-2/(e^2)$	Get the negative reciprocal

Next, substitute the components to the point gradient formula.

$x_1$	$=$	$1/(e^2)$

$y_1$	$=$	$-1$

$m$	$=$	$-2/(e^2)$

$y-$ $y_1$	$=$	$m$ $(x-$ $x_1$ $)$

$y-$ $-1$	$=$	$-2/(e^2)$ $(x-$ $1/(e^2)$ $)$	Substitute known values

$y+1$	$=$	$-2/(e^2)x+2/(e^4)$	Distribute

$y+1$ $-1$	$=$	$-2/(e^2)x+2/(e^4)$ $-1$	Subtract $1$ from both sides

$y$	$=$	$-2/(e^2)x+2/(e^4)-1$

$y=-2/(e^2)x+2/(e^4)-1$

Log and Exponential Tangents and Normals

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Quiz summary

Information

1. Question

Hint

Well Done!

Derivative of $y=ln f(x)$

Point Gradient Formula

2. Question

Hint

Fantastic!

Product Rule with Base “e”

Point Gradient Formula

3. Question

Hint

Nice Job!

Slope of an Exponential Function

4. Question

Hint

Excellent!

Derivative of $y=ln f(x)$

Point Gradient Formula

Quizzes

$m$	$=$	$y’$
	$=$	$e^x$	$y’=e^x$
	$=$	$e^4$	Substitute known values
	$=$	$54.98$	Compute using calculator

$tantheta$	$=$	$m$
$theta$	$=$	$tan^-1$ $(m)$	Reverse the function to get $theta$
$theta$	$=$	$tan^-1$ $(54.98)$	Substitute known values
$theta$	$=$	$89°$	Compute using calculator

Log and Exponential Tangents and Normals

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Quiz summary

Information

1. Question

Hint

Well Done!

Derivative of y=lnf(x)y=ln f(x)

Point Gradient Formula

2. Question

Hint

Fantastic!

Product Rule with Base “e”

Point Gradient Formula

3. Question

Hint

Nice Job!

Slope of an Exponential Function

4. Question

Hint

Excellent!

Derivative of y=lnf(x)y=ln f(x)

Point Gradient Formula

Quizzes

Derivative of $y=ln f(x)$

Derivative of $y=ln f(x)$