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Log and Exponential Tangents and NormalsLog and Exponential Tangents and Normals
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Question 1 of 4
1. Question
Find the equation of the tangent to the curve `y=ln(5x+5)` at the point `x=3`.
Hint
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Well Done!
Incorrect
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
2. Question
Given the function `y=e^(2x)`, find the equation of the tangent line to the point `(2,e^4)`Hint
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Fantastic!
Incorrect
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``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)`°`
Hint
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Nice Job!
Incorrect
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°` -
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Question 4 of 4
4. Question
Find the equation of the normal line that intersects the curve `y=ln sqrtx` at the point `y=-1`.
Hint
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Excellent!
Incorrect
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`