Find the equation of the tangent to the curve `y=x text{arctan}(x)` at the point with coordinates `(1,(pi)/(4))`. Give your answer in the form `y=mx+c` (3 marks)
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Find the equation of the tangent to the curve `y=x text{arctan}(x)` at the point with coordinates `(1,(pi)/(4))`. Give your answer in the form `y=mx+c` (3 marks)
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`y=((2+pi)/4)x-1/2`
`y` | `=xtan^(-1)(x)` | |
`dy/dx` | `=x xx 1/(1+x^2)+tan^(-1)(x)` |
`text{When}\ \ x=1:`
`dy/dx=1/2+tan^(-1)(1)=1/2+pi/4=(2+pi)/4`
`text{Equation of tangent}\ \ m=(2+pi)/4,\ text{through}\ \ (1,(pi)/(4)):`
`y-pi/4` | `=(2+pi)/4 (x-1)` | |
`y` | `=((2+pi)/4)x-(2+pi)/4+pi/4` | |
`y` | `=((2+pi)/4)x-1/2` |
Let `f(x) = tan^(-1) (3x - 6) + pi`.
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a. | `f prime(x)` | `= (d/(dx) (3x – 6))/(1 + (3x – 6)^2)` |
`= 3/(9x^2 – 36x + 37)` |
b. `f″(x) = (3(18x – 36))/(9x^2 – 36x + 37)^2`
`f″(x) = 0\ \ text(when)\ \ 18x – 36 = 0 \ => \ x = 2`
`text(If)\ \ x < 2, 18x – 36 < 0 \ => \ f″(x) < 0`
`text(If)\ \ x > 2, 18x – 36 > 0 \ => \ f″(x) > 0`
`text(S) text(ince)\ \ f″(x)\ \ text(changes sign about)\ \ x = 2,`
`text(a POI exists at)\ \ x = 2`
c. |
Suppose `f(x) = tan(cos^(−1)(x))` and `g(x) = (sqrt(1 - x^2))/x`.
The graph of `y = g(x)` is given.
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i. `f(x) = tan(cos^(−1)(x))`
`f^(′)(x)` | `= −1/sqrt(1 – x^2) · sec^2(cos^(−1)(x))` |
`= −1/sqrt(1 – x^2) · 1/(cos^2(cos^(−1)(x)))` | |
`= −1/(x^2sqrt(1 – x^2))` |
`g(x) = (1 – x^2)^(1/2) · x^(−1)`
`g^(′)(x)` | `= 1/2 · −2x(1 – x^2)^(−1/2) · x^(−1) – (1 – x^2)^(1/2) · x^(−2)` |
`= (−x)/(x sqrt(1 – x^2)) – sqrt(1 – x^2)/(x^2)` | |
`= (−x^2 – sqrt(1 – x^2) sqrt(1 – x^2))/(x^2 sqrt(1 – x^2))` | |
`= (−x^2 – (1 – x^2))/(x^2sqrt(1 – x^2))` | |
`= −1/(x^2sqrt(1 – x^2))` | |
`=f^(′)(x)` |
ii. `f^(′)(x) = g^(′)(x)`
`=> f(x) = g(x) + c`
`text(Find)\ c:`
`f(1)` | `= tan(cos^(−1) 1)` |
`= tan 0` | |
`= 0` |
`g(1) = sqrt(1 – 1)/0 = 0`
`f(1) = g(1) + c`
`:. c = 0`
`:. f(x) = g(x)`
What is the derivative of `tan^(-1)\ x/2`?
A. `1/(2(4 + x^2))`
B. `1/(4 + x^2)`
C. `2/(4 + x^2)`
D. `4/(4 + x^2)`
`C`
`y` | `= tan^(-1)\ x/2` |
`(dy)/(dx)` | `= (1/2)/(1 + (x/2)^2)` |
`= 1/(2(1 + x^2/4))` | |
`= 2/(4 + x^2)` |
`=> C`
Differentiate `3tan^(−1)(2x)`. (2 marks)
`6/(1 + 4x^2)`
`y` | `= 3 tan^-1 (2x)` |
`(dy)/(dx)` | `= 3/(1 + (2x)^2) xx 2` |
`= 6/(1 + 4x^2)` |
Differentiate `tan^(–1)(x^4)` with respect to `x`. (2 marks)
`(4x^3)/(1 + x^8)`
`y` | `= tan^(−1)(x^4)` |
`(dy)/(dx)` | `= 1/(1 + (x^4)^2) xx d/(dx) (x^4)` |
`= (4x^3)/(1 + x^8)` |
Let `f(x) = tan^(-1)(x) + tan^(-1)(1/x)` for `x != 0`.
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i. | `f(x)` | `= tan^(-1) (x) + tan^(-1) (1/x)\ text(for)\ x != 0` |
`f prime (x)` | `= 1/(1 + x^2) + 1/(1 + (1/x)^2) xx d/(dx) (1/x)` | |
`= 1/(1 + x^2) + 1/(1 + 1/(x^2)) xx -1/(x^2)` | ||
`= 1/(1 + x^2)\ – 1/(x^2 + 1)` | ||
`= 0` |
`text(S)text(ince)\ \ f prime (x) = 0`
`=> f(x)\ text(is a constant)`
`text(Substitute)\ \ x = 1\ \ text(into)\ \ f(x)`
`f(1)` | `= tan^(-1) 1 + tan^(-1) (1/1)` |
`= pi/4 + pi/4` | |
`= pi/2` |
`:.\ f(x) = pi/2\ \ text(for)\ \ x > 0\ \ \ …\ text(as required)`
ii. | `text(Given)\ \ f(x)\ \ text(is odd)` |
`f(–x) = -f(x)` |