h(x)=6x−2+7x2h(x)=6x-2+7×2 Find the second derivative of the function.

# h(x)=6x−2+7x2h(x)=6x-2+7×2 Find the second derivative of the function.

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f(x)=(34)x+2,[0,4]f(x)=(34)x+2,[0,4] Find the absolute extrema of the function on the closed interval.

Given: f(x)=(34)x+2,[0,4]f(x)=(34)x+2,[0,4] First find the critical value(s) of the function (if any exist). To find the critical value(s) of the function, set the derivative equal to zero and solve for the…

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f(x)=3−x,[−1,2]f(x)=3-x,[-1,2] Find the absolute extrema of the function on the closed interval.

Given: f(x)=3-x,[-1,2] First find the critical values of the function. To find the critical values of the function (if one exists), set the derivative of the function equal to zero and solve for…

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f(θ)=2sec(θ)+tan(θ),0<θ<2πf(θ)=2sec(θ)+tan(θ),0<θ<2π Find the critical numbers of…

Given the function f(x) =2sec(θ)θ)+tan(θ)θ) in the interval 0<theta<2 pi We have to find the critical points. So let us first take the derivative of the function and evaluate it to…

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h(x)=sin2(x)+cos(x),0<x<2πh(x)=sin2(x)+cos(x),0<x<2π Find the critical numbers of the function.

Given the function h(x)=sin2(x)+cos(x)h(x)=sin2(x)+cos(x) in the interval 0<x<2π0<x<2π We have to find the critical numbers of the function. First take the derivative of the function and equate it to zero. We…

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f(x)=4xx2+1f(x)=4xx2+1 Find the critical numbers of the function.

Given f(x)=4xx2+1f(x)=4xx2+1 Find the derivative using the Quotient Rule. The set the derivative equal to zero and solve for the critical x value(s). f'(x)=(x2+1)(4)−(4x)(2x)x2+1=0f′(x)=(x2+1)(4)-(4x)(2x)x2+1=0…

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g(t)=t4−t−−−−√,t<3g(t)=t4-t,t<3 Find the critical numbers of the function.

You need to evaluate the critical numbers of the function and for this reason, you must differentiate the function with respect to t, using the product and chain rules, such that:

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g(x)=x4−8x2g(x)=x4-8×2 Find the critical numbers of the function.

You need to evaluate the critical numbers of the function and for this reason, you must differentiate the function with respect to x, such that: f'(x)=(x4−8×2)’f′(x)=(x4-8×2)′ f'(x)=4×3−16xf′(x)=4×3-16x You…

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f(x)=x3−3x2f(x)=x3-3×2 Find the critical numbers of the function.

You need to evaluate the critical numbers of the function and for this reason, you must differentiate the function with respect to x, such that: f'(x)=(x3−3×2)’f′(x)=(x3-3×2)′ f'(x)=3×2−6xf′(x)=3×2-6x You need…

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cos(x+y)=xcos(x+y)=x Find dydxdydx by implicit differentiation.

Note:- If y = cos(x) ; then dy/dx = -sin(x) Now, cos(x+y)=xcos(x+y)=x or,−sin(x+y)⋅{1+(dydx)}=1or,-sin(x+y)⋅{1+(dydx)}=1 or,1+(dydx)=−1sin(x+y)or,1+(dydx)=-1sin(x+y) or,1+(dydx)=−cosec(x+y)or,1+(dydx)=-cosec(x+y) or,dydx=−1−cosec(x+y)or,dydx=-1-cosec(x+y)

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xsin(y)=ycos(x)xsin(y)=ycos(x) Find dydxdydx by implicit differentiation.

Note:- 1) If y = sinx; then dy/dx = cosx 2) If y = cos(x) ; then dy/dx = -sinx Now, x⋅sin(y)=y⋅cos(x)x⋅sin(y)=y⋅cos(x) or,x⋅cosy⋅(dydx)+sin(y)=−y⋅sin(x)+cos(x)⋅(dydx)or,x⋅cosy⋅(dydx)+sin(y)=-y⋅sin(x)+cos(x)⋅(dydx)

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xy−−√=x−4yxy=x-4y Find dydxdydx by implicit differentiation.

xy−−√=x−4yxy=x-4y Differentiating both sides with respect to x, (12)(xy)−12ddx(xy)=1−4y'(12)(xy)-12ddx(xy)=1-4y′ 12xy−−√(xy’+y)=1−4y’12xy(xy′+y)=1-4y′ xy’+y=2xy−−√(1−4y’)xy′+y=2xy(1-4y′) xy’+y=2xy−−√−8xy−−√⋅y’xy′+y=2xy-8xy⋅y′…

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(x3)y−x(y3)=4(x3)y-x(y3)=4 Find dydxdydx by implicit differentiation.

(x3)⋅y−x⋅(y3)=4(x3)⋅y-x⋅(y3)=4 differentiating 3(x2)⋅y+(x3)⋅(dydx)−(y3)−3x(y2)⋅(dydx)=03(x2)⋅y+(x3)⋅(dydx)-(y3)-3x(y2)⋅(dydx)=0 or,(dydx)⋅[(x3)−3x(y2)]=(y3)−3(x2)yor,(dydx)⋅[(x3)-3x(y2)]=(y3)-3(x2)y or,dydx=(y3)−3(x2)y(x3)−3x(y2)or,dydx=(y3)-3(x2)y(x3)-3x(y2)

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x2+4xy−y3=6×2+4xy-y3=6 Find dydxdydx by implicit differentiation.

(x2)+4xy−(y3)=6(x2)+4xy-(y3)=6 differentiating 2x+4y+4x(dydx)−3(y2)⋅(dydx)=02x+4y+4x(dydx)-3(y2)⋅(dydx)=0 or,2(x+2y)=(dydx)⋅[3(y2)−4x]or,2(x+2y)=(dydx)⋅[3(y2)-4x] or,dydx=2(x+2y)3(y2)−4xor,dydx=2(x+2y)3(y2)-4x

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x2+y2=64×2+y2=64 Find dydxdydx by implicit differentiation.

(x2)+(y2)=64(x2)+(y2)=64 Differentiating 2x+2y(dydx)=02x+2y(dydx)=0 or,x+y(dydx)=0or,x+y(dydx)=0 or,dydx=−xyor,dydx=-xy

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y=(sin(x))2y=(sin(x))2 Find the second derivative of the function.

Note:- If y = sinx ; then dy/dx = cosx If y = cosx ; then dy/dx = -sinx 2sinx*cosx = sin(2x) Now, y=(sinx)2y=(sinx)2 or,y=sin2xor,y=sin2xdifferentiating y’=2sinx⋅cosxy′=2sinx⋅cosx or,y’=sin(2x)or,y′=sin(2x)…

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f(x)=cot(x)f(x)=cot(x) Find the second derivative of the function.

Note:- 1) If y = cotx ; then dy/dx = −cosec2(x)-cosec2(x)  2) If y = cosecx ; then dy/dx = -cosecx*cotx Now, f(x)=y=cotxf(x)=y=cotx differentiating f'(x)=y’=−cosec2(x)f′(x)=y′=-cosec2(x) differentiating

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y=15x+1y=15x+1 Find the second derivative of the function.

y=15x+1=(5x+1)−1y=15x+1=(5x+1)-1  d⇔erentiat∈gd⇔erentiat∈g y’=−1{(5x+1)−2}⋅5y′=-1{(5x+1)-2}⋅5 or,y’=−5(5x+1)−2or,y′=-5(5x+1)-2 Differentiating againg w.r.t ‘x’ we get y”=10⋅5⋅(5x+1)−3y′′=10⋅5⋅(5x+1)-3 or,y”=50(5x+1)3or,y′′=50(5x+1)3

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y=(8x+5)3y=(8x+5)3 Find the second derivative of the function.

y=(8x+5)3y=(8x+5)3 y’=3⋅8⋅(8x+5)2y′=3⋅8⋅(8x+5)2 y’=24⋅(8x+5)2y′=24⋅(8x+5)2 y”=24⋅2⋅8⋅(8x+5)y′′=24⋅2⋅8⋅(8x+5) or,y”=384⋅(8x+5)or,y′′=384⋅(8x+5)

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y=csc(3x)+cot(3x),((π6),1)y=csc(3x)+cot(3x),((π6),1) Find and evaluate the derivative of the function at the…

y=cosec(3x)+cot(3x)y=cosec(3x)+cot(3x) y’=−3⋅cosec(3x)⋅cot(3x)−3⋅cosec2(3x)y′=-3⋅cosec(3x)⋅cot(3x)-3⋅cosec2(3x) Putting x = pi/6 we get y’ = −3⋅cosex(π2)⋅cot(π2)−3⋅cosec2(π2)-3⋅cosex(π2)⋅cot(π2)-3⋅cosec2(π2) or,y’=−3⋅1⋅0−3⋅1or,y′=-3⋅1⋅0-3⋅1 or,y’=−3or,y′=-3

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y=(12)csc(2x),(π412)y=(12)csc(2x),(π412) Find and evaluate the derivative of the function at the…

y=(12)⋅cosec(2x)y=(12)⋅cosec(2x) y’=−(12)⋅2⋅cosec(2x)⋅cot(2x)y′=-(12)⋅2⋅cosec(2x)⋅cot(2x)Putting x = pi/4 y’=−1⋅cosec(π2)⋅cot(π2)=−1⋅1⋅0=0y′=-1⋅cosec(π2)⋅cot(π2)=-1⋅1⋅0=0

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f(x)=3x+14x−3,(4,1)f(x)=3x+14x-3,(4,1) Find and evaluate the derivative of the function at the…

f(x)=3x+14x−3f(x)=3x+14x-3 f'(x)=3⋅(4x−3)−4⋅(3x+1)(4x−3)2f′(x)=3⋅(4x-3)-4⋅(3x+1)(4x-3)2 or,f'(x)=−13(4x−3)2or,f′(x)=-13(4x-3)2 or,f'(4)=−13(4⋅4−3)2or,f′(4)=-13(4⋅4-3)2 or,f'(4)=−13(13)2or,f′(4)=-13(13)2 or,f'(4)=−113or,f′(4)=-113

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f(x)=4×2+1,(−1,2)f(x)=4×2+1,(-1,2) Find and evaluate the derivative of the function at the given…

f(x)=4×2+1f(x)=4×2+1 or,f(x)=4⋅(x2+1)−1or,f(x)=4⋅(x2+1)-1 thus,f'(x)=−4⋅(2x)⋅(x2+1)−2thus,f′(x)=-4⋅(2x)⋅(x2+1)-2 or,f'(x)=−8x(x2+1)2or,f′(x)=-8x(x2+1)2 or,f'(−1)=−8⋅−1{(−1)2+1}2or,f′(-1)=-8⋅-1{(-1)2+1}2 or,f'(x)=84=2or,f′(x)=84=2

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f(x)=x2−1−−−−−√3,(3,2)f(x)=x2-13,(3,2) Find and evaluate the derivative of the function at the…

f(x)=x2−1−−−−−√3f(x)=x2-13  f'(x)=(13)(x2−1)13−1(2x)f′(x)=(13)(x2-1)13-1(2x) f'(x)=2×3(x2−1)23f′(x)=2×3(x2-1)23 Therefore the derivative at the point(3,2) can be obtained by plugging in the value of x in the f'(x). f'(x)…

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f(x)=1−x3−−−−−√,(−2,3)f(x)=1-x3,(-2,3) Find and evaluate the derivative of the function at the given…

f(x)=1−x3−−−−−√f(x)=1-x3 f'(x)=(12)(1−x3)12−1(−3×2)f′(x)=(12)(1-x3)12-1(-3×2) f'(x)=−3×221−x3−−−−−√f′(x)=-3×221-x3 Therefore the derivative of the function at the point (-2,3) can be obtained by plugging in the value of of x…

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h(x)=(x+5×2+3)2h(x)=(x+5×2+3)2 Find the derivative of the function.

Given: h(x)=(x+5×2+3)2h(x)=(x+5×2+3)2 To find the derivative of the function use the Quotient Rule within the Chain Rule. h(x)=2(x+5×2+3)[(x2+3)(1)−(x+5)(2x)(x+3)2]h(x)=2(x+5×2+3)[(x2+3)(1)-(x+5)(2x)(x+3)2]…

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f(x)=3xx2+1−−−−−√f(x)=3xx2+1 Find the derivative of the function.

f(x)=3xx2+1−−−−−√f(x)=3xx2+1 Using the quotient rule of the derivative, f'(x)=3(x2+1−−−−−√−xddx(x2+1−−−−−√)(x2+1−−−−−√)2)f′(x)=3(x2+1-xddx(x2+1)(x2+1)2) f'(x)=3⎛⎝x2+1−−−−−√−x(12)(x2+1)−122xx2+1⎞⎠f′(x)=3(x2+1-x(12)(x2+1)-122xx2+1)…

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f(s)=(s2−1)52(s3+5)f(s)=(s2-1)52(s3+5) Find the derivative of the function.

f(s)=(s2−1)52(s3+5)f(s)=(s2-1)52(s3+5) Using the product rule for the derivative, f'(s)=(s2−1)52dds(s3+5)+(s3+5)dds(s2−1)52f′(s)=(s2-1)52dds(s3+5)+(s3+5)dds(s2-1)52 f'(s)=(s2−1)52(3s2)+(s3+5)(52)(s2−1)(52)−1(2s)f′(s)=(s2-1)52(3s2)+(s3+5)(52)(s2-1)(52)-1(2s)…

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y=x(6x+1)5y=x(6x+1)5 Find the derivative of the function.

You need to differentiate the function with respect to x, using the product rule and chain rule, such that: y’=x’⋅(6x+1)5+x⋅((6x+1)5)’y′=x′⋅(6x+1)5+x⋅((6x+1)5)′ y’=1⋅(6x+1)5+x⋅5⋅(6x+1)4⋅(6x+1)’y′=1⋅(6x+1)5+x⋅5⋅(6x+1)4⋅(6x+1)′

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y=(sec(x))77−(sec(x))55y=(sec(x))77-(sec(x))55 Find the derivative of the function.

Given: y=(sec(x))77−(sec(x))55y=(sec(x))77-(sec(x))55 y’=7sec(x)67sec(x)tan(x)−5sec(x)45sec(x)tan(x)y′=7sec(x)67sec(x)tan(x)-5sec(x)45sec(x)tan(x) y’=(sec(x))7tan(x)−(sec(x))5tan(x)y′=(sec(x))7tan(x)-(sec(x))5tan(x) y’=(sec(x))5tan(x)[(sec(x))2−1]y′=(sec(x))5tan(x)[(sec(x))2-1]…

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y=x2−sin(2x)4y=x2-sin(2x)4 Find the derivative of the function.

Note:- 1) If y = sin(x) ; then dy/dx = cos(x) 2) If y = x^n ; then dy/dx = n*x^(n-1) ; where n = constant Now, y=(x2)−{sin(2x)4}y=(x2)-{sin(2x)4} dydx=y’=(12)−2cos(2x)4dydx=y′=(12)-2cos(2x)4

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y=1−cos(2x)+2(cos(x))2y=1-cos(2x)+2(cos(x))2 Find the derivative of the function.

y = 1- cos(2x) + 2(cos2x)(cos2x) dydx=y’=2⋅sin(2x)−4⋅cosx⋅sinxdydx=y′=2⋅sin(2x)-4⋅cosx⋅sinxor,y’=2sin(2x)−2sin(2x)=0or,y′=2sin(2x)-2sin(2x)=0 note: 2sinx⋅cosx=sin2x2sinx⋅cosx=sin2x

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y=5cos(9x+1)y=5cos(9x+1) Find the derivative of the function.

Note:- If y = cos(ax) ; then dy/dx = -a*sin(ax) Now, y=5cos(9x+1)y=5cos(9x+1) dydx=y’=−5⋅(sin(9x+1))⋅9dydx=y′=-5⋅(sin(9x+1))⋅9 or,dydx=y’=−45sin(9x+1)or,dydx=y′=-45sin(9x+1)

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f(x)=1(5x+1)2f(x)=1(5x+1)2 Find the derivative of the function.

Note :- 1) If y = x^n ; where n = constant, then dy/dx = n*(x^(n-1)) 2) If y = n*x ; where n = constant ; then dy/dx = n Now, y=1(5x+1)2y=1(5x+1)2 or,y=(5x+1)−2or,y=(5x+1)-2

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y=1×2+4y=1×2+4 Find the derivative of the function.

Note:- If y = x^n ; where n = constant, then dy/dx = n*x^(n-1) Now, y=1(x2)+4y=1(x2)+4 Thus,y={(x2)+4}−1Thus,y={(x2)+4}-1 or,y’=−1{{(x2)+4}−2}⋅(2x)or,y′=-1{{(x2)+4}-2}⋅(2x) or,y’=−2x{(x2)+4}2or,y′=-2x{(x2)+4}2

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y=(x2−6)3y=(x2-6)3 Find the derivative of the function.

Note:- If y = x^n ; where n = constant ; then dy/dx = n*{x^(n-1)} Now, y=(x2−6)3y=(x2-6)3 thus,dydx=y’=3⋅{(x2−6)2}⋅(2x)thus,dydx=y′=3⋅{(x2-6)2}⋅(2x) or,dydx=y’=6x⋅(x2−6)2or,dydx=y′=6x⋅(x2-6)2

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y=(7x+3)4y=(7x+3)4 Find the derivative of the function.

Note:- If y = x^n ; where n = constant ; then dy/dx = n*{x^(n-1)} Now, y = (7x+3)4(7x+3)4 thus,dydx=y’=4{(7x+3)3}⋅7thus,dydx=y′=4{(7x+3)3}⋅7 or,dydx=y’=28(7x+3)3or,dydx=y′=28(7x+3)3

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h(t)=10cos(t)−15sin(t)h(t)=10cos(t)-15sin(t) Find the second derivative of the function.

Note:- If y = cos(ax) ; then dy/dx = -a*sin(ax) If y = sin(ax) ; then dy/dx = a*cos(ax) ; where ‘a’ = constant Now, h(t)=10cos(t)−15sin(t)h(t)=10cos(t)-15sin(t) h'(t)=−10sin(t)−15cos(t)h′(t)=-10sin(t)-15cos(t)

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f(θ)=3tan(θ)f(θ)=3tan(θ) Find the second derivative of the function.

Note:- If y = tanx ; then dy/dx = sec^2(x) If y = sec(x) ; then dy/dx = sec(x)*tan(x) Now, f(θ)=3tan(θ)f(θ)=3tan(θ) f'(θ)=3sec2(θ)f′(θ)=3sec2(θ) f”(θ)=3⋅2sec(θ)⋅sec(θ)⋅tan(θ)f′′(θ)=3⋅2sec(θ)⋅sec(θ)⋅tan(θ)…

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f(x)=20x−−√5f(x)=20×5 Find the second derivative of the function.

Note:- If y = x^n ; where n = constant ; then dy/dx = n*{x^(n-1)} Now, y=20⋅x15y=20⋅x15 y’=20⋅(15)⋅x(15)−1y′=20⋅(15)⋅x(15)-1 or,y’=4⋅x−45or,y′=4⋅x-45 thus,y”=4⋅(−45)⋅x(−45)−1thus,y′′=4⋅(-45)⋅x(-45)-1

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f(x)=15x52f(x)=15×52 Find the second derivative of the function.

Note:- If y = x^n ; where n = constant ; then dy/dx = n*{x^(n-1)} Now, y=15⋅x52y=15⋅x52 y’=15⋅(52)⋅x(52)−1y′=15⋅(52)⋅x(52)-1 or,y’=(752)⋅x32or,y′=(752)⋅x32 thus,y”=(752)⋅(32)⋅x(32)−1thus,y′′=(752)⋅(32)⋅x(32)-1

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h(x)=6x−2+7x2h(x)=6x-2+7×2 Find the second derivative of the function.

Note:- If y = x^n ; where n = constant ; then dy/dx = n*x^(n-1) Now, h(x)=6x−2+7x2h(x)=6x-2+7×2 h'(x)=−12x−3+14xh′(x)=-12x-3+14x h”(x)=−12⋅(−3)⋅x−4+14h′′(x)=-12⋅(-3)⋅x-4+14 or,h”(x)=36x−4+14or,h′′(x)=36x-4+14

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g(t)=−8t3−5t+12g(t)=-8t3-5t+12 Find the second derivative of the function.

Note:- If y = x^n ; where n = constant ; then dy/dx = n*x^(n-1) Now, g(t)=−8t3−5t+12g(t)=-8t3-5t+12 g'(t)=−8⋅3⋅t2−5+0g′(t)=-8⋅3⋅t2-5+0 or,g'(t)=−24t2−5or,g′(t)=-24t2-5 Thus,g”(t)=−24⋅2⋅t1−0Thus,g′′(t)=-24⋅2⋅t1-0 or,g”(t)=−48tor,g′′(t)=-48t

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f(x)=1+cos(x)1−cos(x),((π2),1)f(x)=1+cos(x)1-cos(x),((π2),1) Find an equation of the tangent line to the…

Hello! f(x)=1+cos(x)1−cos(x).f(x)=1+cos(x)1-cos(x). First, let’s check that f(π2)=1f(π2)=1 : cos(π2)=0cos(π2)=0 and (1+0)/(1-0) = 1. Then remember that1+cos(x)=2⋅[cos(x2)]21+cos(x)=2⋅[cos(x2)]2 and 1−cos(x)=2⋅[sin(x2)]21-cos(x)=2⋅[sin(x2)]2 ….

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f(x)=x+1x−1,((12),−3)f(x)=x+1x-1,((12),-3) Find an equation of the tangent line to the graph of…

f(x)=x+1x−1f(x)=x+1x-1 f'(x)=(x−1)⋅1−(x+1)⋅1(x−1)2f′(x)=(x-1)⋅1-(x+1)⋅1(x-1)2 or,f'(x)=−2(x−1)2or,f′(x)=-2(x-1)2 Now, slope of the tangent at point ((1/2),-3) = f'(1/2) = -8 Thus, equation of the tangent to the curve at the…

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f(x)=(x−4)(x2+6x−1),(0,4)f(x)=(x-4)(x2+6x-1),(0,4) Find an equation of the tangent line to the graph of…

Given: f(x)=(x−4)(x2+6x−1),(0,4)f(x)=(x-4)(x2+6x-1),(0,4) Find the derivative of the function using the Product Rule. Then plug in the given x value into the derivative function to calculate the slope….

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• MATH

f(x)=(x+2)(x2+5),(−1,6)f(x)=(x+2)(x2+5),(-1,6) Find an equation of the tangent line to the graph of ff…

Given: f(x)=(x+2)(x2+5),(−1,6)f(x)=(x+2)(x2+5),(-1,6) Find the derivative using the Quotient Rule. Then substitute in the given x value into the f'(x) equation to calculate the slope. f'(x)=(x+2)(2x)+(x2+5)(1)f′(x)=(x+2)(2x)+(x2+5)(1)…

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• MATH

g(x)=3xsin(x)+(x2)cos(x)g(x)=3xsin(x)+(x2)cos(x) Use the Product Rule or the Quotient Rule to find the…

You need to evaluate the derivative of the given function, using the product tule for the products 3x⋅sinx3x⋅sinx and x2⋅cosxx2⋅cosx , such that:

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• MATH

y=xcos(x)−sin(x)y=xcos(x)-sin(x) Use the Product Rule or the Quotient Rule to find the derivative of…

You need to evaluate the derivative of the given function, using the product rule for the product x*cos x, such that: f'(x)=((x)'(cosx)+(x)(cosx)’)−(sinx)’f′(x)=((x)′(cosx)+(x)(cosx)′)-(sinx)′

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• MATH

y=2x−(x2)tan(x)y=2x-(x2)tan(x) Use the Product Rule or the Quotient Rule to find the derivative of…

y=2x−x2tan(x)y=2x-x2tan(x) Now to evaluate the derivative , let us first apply the Sum/Difference rule i.e. (f±g)’=f’±g'(f±g)′=f′±g′ y’=ddx(2x)−ddx(x2tan(x))y′=ddx(2x)-ddx(x2tan(x)) y’=2−ddx(x2tan(x))y′=2-ddx(x2tan(x)) Now applying the product rule…

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• MATH

y=3(x2)sec(x)y=3(x2)sec(x) Use the Product Rule or the Quotient Rule to find the derivative of the…

You need to evaluate the derivative of the given function, using the product tule for the product 3×2⋅secx3x2⋅secx , such that: f'(x)=(3×2)’⋅(secx)+3×2⋅(secx)’f′(x)=(3×2)′⋅(secx)+3×2⋅(secx)′

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