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Applications of Differentiation>
Applications of DifferentiationApplications of Differentiation
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Question 1 of 3
1. Question
Given that `x=(pi)/12`, find the slope (`m`) of the normal line`y=sin3x`Hint
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Derivatives of Trigonometric Functions
`y'(\text(sin))=\text(cos)``y'(\text(cos))=-\text(sin)``y'(\text(tan))=\text(sec)^2`First, find the derivative of the equation`y` `=` `\text(sin) 3x` `y’` `=` `\text(cos) 3x*3` Derive `\text(sin) 3x` `=` `3 \text(cos) 3x` Next, substitute the value of `x` and get the slope of the tangent`m` `=` `3 \text(cos) 3x` `=` `3 \text(cos) 3*((pi)/12)` Substitute `x=(pi)/12` `=` `3 \text(cos) (pi)/4` Simplify `=` `3*1/(sqrt2)` `\text(cos) (pi)/4=1/(sqrt2)` `=` `3/(sqrt2)` Finally, to find the normal line, get the negative reciprocal of the tangent, which is `3/(sqrt2)``3/(sqrt2)` `=` `-(sqrt2)/3` Get the negative reciprocal `m=-(sqrt2)/3` -
Question 2 of 3
2. Question
Given that `x=(pi)/2`, find the slope (`m`) of the tangent line`y=tan(x/2)`- `m=` (1)
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Derivatives of Trigonometric Functions
`y'(\text(sin))=\text(cos)``y'(\text(cos))=-\text(sin)``y'(\text(tan))=\text(sec)^2`First, find the derivative of the equation`y` `=` `\text(tan) x/2` `y’` `=` `(\text(sec) x/2)^2*1/2` Derive `\text(tan) x/2` `=` `1/2(\text(sec) x/2)^2` Finally, substitute the value of `x` and get the slope of the tangent`m` `=` `1/2(\text(sec) x/2)^2` `=` `1/2(\text(sec) ((pi)/2)/2)^2` Substitute `x=(pi)/1` `=` `1/2(\text(sec) (pi)/4)^2` Simplify `=` `1/2(sqrt2)^2` `\text(sec) (pi)/4=sqrt2` `=` `1/2*2` `=` `1` `m=1` -
Question 3 of 3
3. Question
Find the equation of the tangent to the curve `y=cos2x` at the point `x=pi/4`Hint
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Derivatives of Trigonometric Functions
`y'(\text(sin))=\text(cos)``y'(\text(cos))=-\text(sin)``y'(\text(tan))=\text(sec)^2`Point Gradient Formula
`y-``y_1``=``m``(x-``x_1``)`First, find the derivative of the equation`y` `=` `\text(cos) 2x` `y’` `=` `-\text(sin) 2x^2*2` Derive `\text(cos) 2x` `=` `-2 \text(sin) 2x` Next, substitute the value of `x` to the derived function and get the slope of the tangent`-2 \text(sin) 2x` `=` `-2 \text(sin) 2(pi/4)` Substitute `x=pi/4` `=` `-2 \text(sin) pi/2` Simplify `=` `-2*1` `\text(sin) (pi)/2=1` `=` `-2` Next, substitute the value of `x` to the original expression and get value of `y``y` `=` `\text(cos) 2x` `y` `=` `\text(cos) 2(pi/4)` Substitute `x=pi/4` `=` `\text(cos) pi/2` Simplify `=` `0` `\text(cos) (pi)/2=0` Finally, substitute the known values to the slope-gradient formula to get the equation`x_1` `=` `pi/4` `y_1` `=` `0` `m` `=` `-2` `y-``y_1` `=` `m``(x-``x_1``)` `y-``0` `=` $$\color{#007986}{-2}\left(x-\color{#e65021}{\frac{\pi}{4}}\right)$$ Substitute known values `y` `=` `-2x+pi/2` Simplify `y=-2x+pi/2`
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