y=xsin(x)+cos(x)y=xsin(x)+cos(x) Find the derivative of the trigonometric function.

# y=xsin(x)+cos(x)y=xsin(x)+cos(x) Find the derivative of the trigonometric function.

• MATH

f(x)=x2x2+1f(x)=x2x2+1 Determine the point(s) at which the graph of the function has a…

Given: f(x)=x2x2+1f(x)=x2x2+1   Find the derivative of the function using the Quotient Rule. Set the derivative equal zero to find the critical x value(s)….

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f(x)=2x−1x2f(x)=2x-1×2 Determine the point(s) at which the graph of the function has a…

Given f(x)=2x−1x2f(x)=2x-1×2 Find the derivative of the function using the Quotient Rule. Set the derivative equal to zero to find the critical x value(s). When the derivative is equal to zero the…

• MATH

f(x)=sec(x),(π3,2)f(x)=sec(x),(π3,2) (a) Find an equation of the tangent line to the graph of f at the…

You need to evaluate the equation of the tangent line to the curvef(x)=secxf(x)=secx , t the point (π3,2),(π3,2), using the following formula, such that: f(x)−f(π3)=f'(π3)(x−π3)f(x)-f(π3)=f′(π3)(x-π3)…

• MATH

f(x)=tan(x),(π4,1)f(x)=tan(x),(π4,1) (a) Find an equation of the tangent line to the graph of f at the…

You need to evaluate the equation of the tangent line to the curve f(x)=tanxf(x)=tanx , t the point (π4,1),(π4,1), using the following formula, such that: f(x)−f(π4)=f'(π4)(x−π4)f(x)-f(π4)=f′(π4)(x-π4)…

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f(x)=x+3x−3,(4,7)f(x)=x+3x-3,(4,7) (a) Find an equation of the tangent line to the graph of f at…

You need to evaluate the equation of the tangent line at (4,7), using the formula: f(x)−f(4)=f'(4)(x−4)f(x)-f(4)=f′(4)(x-4) Notice that f(4) = 7. You need to evaluate f'(x), using the quotient rule, such…

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f(x)=xx+4,(−5,5)f(x)=xx+4,(-5,5) (a) Find an equation of the tangent line to the graph of f at the…

You need to evaluate the equation of the tangent line at (-5,5), using the formula: f(x)−f(−5)=f'(−5)(x+5)f(x)-f(-5)=f′(-5)(x+5) Notice that f(-5) = 5. You need to evaluate f'(x), using the quotient rule, such…

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f(x)=(x−2)(x2+4),(1,−5)f(x)=(x-2)(x2+4),(1,-5) (a) Find an equation of the tangent line to the graph of f…

The function f(x)=x−2×2+4f(x)=x-2×2+4 The slope of the tangent at point x = a is f'(a). f'(x)=(x−2)’⋅(x2+4)−(x−2)(x2+4)'(x2+4)2f′(x)=(x-2)′⋅(x2+4)-(x-2)(x2+4)′(x2+4)2 = (x2+4)−(x−2)(2x)(x2+4)2(x2+4)-(x-2)(2x)(x2+4)2 =

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f(x)=(x3+4x−1)(x−2),(1,−4)f(x)=(x3+4x-1)(x-2),(1,-4) (a) Find an equation of the tangent line to the graph…

The function f(x)=(x3+4x−1)⋅(x−2)f(x)=(x3+4x-1)⋅(x-2) . The slope of the tangent at point x = a, is given by f'(a). f'(x)=((x3+4x−1)⋅(x−2))’f′(x)=((x3+4x-1)⋅(x-2))′ = (x3+4x−1)’⋅(x−2)+(x3+4x−1)⋅(x−2)'(x3+4x-1)′⋅(x-2)+(x3+4x-1)⋅(x-2)′ =

• MATH

f(x)=sin(x)(sin(x)+cos(x)),(π4,1)f(x)=sin(x)(sin(x)+cos(x)),(π4,1) Evaluate the derivative of the function at the…

The function f(x) = sin x*(sin x + cos x). Use the product rule to determine the derivative of f(x). f'(x) = (sin x)’*(sin x + cos x) + sin x*(sin x + cos x)’ f'(x) = (cos x)*(sin x + cos x) +…

• MATH

h(t)=sec(t)t,(π,−1π)h(t)=sec(t)t,(π,-1π) Evaluate the derivative of the function at the given point….

You need to evaluate the equation of the tangent line to the curve f(t)=secttf(t)=sectt , at the point (π,−1π),(π,-1π), using the following formula, such that: f(t)−f(π)=f'(π)(t−π)f(t)-f(π)=f′(π)(t-π) Notice…

• MATH

f(x)=tan(x)cot(x),(1,1)f(x)=tan(x)cot(x),(1,1) Evaluate the derivative of the function at the given point. Use…

You need to evaluate the equation of the tangent line to the curve f(x) = tan x*cot x, t the point (1, 1), using the following formula, such that: f(x)−f(1)=f'(1)(x−1)f(x)-f(1)=f′(1)(x-1) Notice that f(11 ) =…

• MATH

y=1+csc(x)1−csc(x),(π6,−3)y=1+csc(x)1-csc(x),(π6,-3) Evaluate the derivative of the function at the…

y=1+csc(x)1−csc(x)y=1+csc(x)1-csc(x) differentiating by applying the quotient rule, y’=(1−csc(x))ddx(1+csc(x))−(1+csc(x))ddx(1−csc(x))(1−csc(x))2y′=(1-csc(x))ddx(1+csc(x))-(1+csc(x))ddx(1-csc(x))(1-csc(x))2…

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h(θ)=5θsec(θ)+θtan(θ)h(θ)=5θsec(θ)+θtan(θ) Find the derivative of the trigonometric…

h(θ)=5(θ)sec(θ)+θtan(θ)h(θ)=5(θ)sec(θ)+θtan(θ) h'(θ)=5sec(θ)+5θsec(θ)tan(θ)+tan(θ)+θsec2(θ)h′(θ)=5sec(θ)+5θsec(θ)tan(θ)+tan(θ)+θsec2(θ)

• MATH

y=2xsin(x)+(x2)cos(x)y=2xsin(x)+(x2)cos(x) Find the derivative of the trigonometric function.

y=2xsin(x)+(x2)cos(x)y=2xsin(x)+(x2)cos(x) y’=dydx=2sin(x)+2xcos(x)+2xcos(x)−(x2)sin(x)y′=dydx=2sin(x)+2xcos(x)+2xcos(x)-(x2)sin(x) y’=dydx=2sin(x)+4xcos(x)−(x2)sin(x)y′=dydx=2sin(x)+4xcos(x)-(x2)sin(x)

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f(x)=sin(x)cos(x)f(x)=sin(x)cos(x) Find the derivative of the trigonometric function.

f(x)=sin(x)cos(x)=(12)sin(2x)f(x)=sin(x)cos(x)=(12)sin(2x) f'(x)=2⋅(12)⋅cos(2x)=cos(2x)f′(x)=2⋅(12)⋅cos(2x)=cos(2x) Note:- 2sin(x)cos(x) = sin(2x)

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f(x)=(x2)tan(x)f(x)=(x2)tan(x) Find the derivative of the trigonometric function.

Note:- 1) If y = tanx ; then dy/dx = sec^2(x) 2) If y = x^n ; where ‘n’ = constant ; then dy/dx = n*x^(n-1) 3) If y = u*v ; where ‘u’ & ‘v’ are functions of ‘x’, then dy/dx = uv’ + vu’ Now,…

• MATH

y=xsin(x)+cos(x)y=xsin(x)+cos(x) Find the derivative of the trigonometric function.

Note:- 1) If y = sinx ; then dy/dx = cosx 2) If y = cosx ; then dy/dx = -sinx 3) If y = u*v; where ‘u’ & ‘v’ are functions of ‘x’ ; then dy/dx = uv’ + vu’ Now, y=xsinx+cosxy=xsinx+cosx

• MATH

y=−csc(x)−sin(x)y=-csc(x)-sin(x) Find the derivative of the trigonometric function.

Note:- 1) If y = sinx ; then dy/dx = cosx 2) If y = cosecx ; then dy/dx = -cosecx*cotx Now, y=−cosec(x)−sin(x)y=-cosec(x)-sin(x) dydx=y’=cosec(x)⋅cot(x)−cos(x)dydx=y′=cosec(x)⋅cot(x)-cos(x)

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y=sec(x)xy=sec(x)x Find the derivative of the trigonometric function.

Note:- 1) If y = secx ; then dy/dx = secx*tanx 2) If y = x^n ; then dy/dx = n*x^(n-1) ; where n = constant 3) If y = u/v ; where ‘u’ & ‘v’ are functions of ‘x’ , then dy/dx = [vu’-uv’]/(v^2)…

• MATH

y=3(1−sin(x))2cos(x)y=3(1-sin(x))2cos(x) Find the derivative of the trigonometric function.

You need to evaluate the derivative of the given function, using the quotient rule, such that: y’=(3−3sinx)’⋅(2cosx)−(3−3sinx)⋅(2cosx)’4cos2xy′=(3-3sinx)′⋅(2cosx)-(3-3sinx)⋅(2cosx)′4cos2x

• MATH

h(x)=1x−12sec(x)h(x)=1x-12sec(x) Find the derivative of the trigonometric function.

Note:- 1) If y = secx ; then dy/dx = sec(x)*tan(x) 2) If y = x^n ; where n = constant ; then dy/dx = n*x^(n-1) Now, y=1x−12sec(x)y=1x-12sec(x) y’=dydx=−1×2−12sec(x)⋅tan(x)y′=dydx=-1×2-12sec(x)⋅tan(x)

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g(t)=t√4+6csc(t)g(t)=t4+6csc(t) Find the derivative of the trigonometric function.

g(t)=t14+6cosec(t)g(t)=t14+6cosec(t) g'(t)=(14)t−34−6cosec(t)cot(t)g′(t)=(14)t-34-6cosec(t)cot(t) Note:- 1) If y = x^n ; where n = constant ; then dy/dx = n*x^(n-1) 2) If y = cosec(x) ; then dy/dx = -cosec(x)*cot(x)

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y=x+cot(x)y=x+cot(x) Find the derivative of the trigonometric function.

Note:- 1) If y = cotx ; then dy/dx = -cosec^2(x) 2) If y = x^n ; where ‘n’ = constant ; then dy/dx = n*x^(n-1) Now, y=x+cot(x)y=x+cot(x) dydx=y’=1−cosec2(x)dydx=y′=1-cosec2(x) or,dydx=−[cosec2(x)−1]or,dydx=-[cosec2(x)-1]…

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f(x)=−x+tan(x)f(x)=-x+tan(x) Find the derivative of the trigonometric function.

Note:- 1) If y = tanx ; then dy/dx = sec^2(x) 2) If y = x^n ; then dy/dx = n*x^(n-1) Now, f(x)=−x+tanxf(x)=-x+tanx f'(x)=−1+sec2(x)f′(x)=-1+sec2(x)

• MATH

f(x)=sin(x)x3f(x)=sin(x)x3 Find the derivative of the trigonometric function.

Note:- 1) If y = sinx ; then dy/dx = cosx 2) If y = x^n ; where n = constant ; then dy/dx = n*x^(n-1) 3) If y = u/v ; where ‘u’ & ‘v’ are functions of ‘x’ ; then dy/dx = [vu’ – uv’]/(v^2) Now,…

• MATH

f(t)=cos(t)tf(t)=cos(t)t Find the derivative of the trigonometric function.

Note:- 1) If y = cosx ; then dy/dx = -sinx 2) If y = u/v ; where ‘u’ & ‘v’ are functions of x or constants ; then dy/dx = [vu’-uv’]/(v^2) Now, f(t)=costtf(t)=costt f'(t)=−t⋅sint−costt2f′(t)=-t⋅sint-costt2…

• MATH

f(θ)=(θ+1)cos(θ)f(θ)=(θ+1)cos(θ) Find the derivative of the trigonometric function.

Note:- if y = cos(x) ; then dy/dx = -sinx Now,  f(θ)={(θ)+1}cos(θ)f(θ)={(θ)+1}cos(θ)thus, f'(θ)=cos(θ)−{(θ)+1}sin(θ)f′(θ)=cos(θ)-{(θ)+1}sin(θ)

• MATH

f(t)=(t2)sin(t)f(t)=(t2)sin(t) Find the derivative of the trigonometric function.

f(t)=(t2)sin(t)f(t)=(t2)sin(t) f'(t)=2tsin(t)+(t2)cos(t)f′(t)=2tsin(t)+(t2)cos(t) or,f'(t)=t[2sin(t)+tcos(t)]or,f′(t)=t[2sin(t)+tcos(t)]

• MATH

f(x)=c2−x2c2+x2f(x)=c2-x2c2+x2 Find the derivative of the algebraic function, c is a…

f(x)=c2−x2c2+x2f(x)=c2-x2c2+x2 f'(x)=−2x(c2+x2)−2x(c2−x2)(c2+x2)2f′(x)=-2x(c2+x2)-2x(c2-x2)(c2+x2)2 or,f'(x)=−4xc2(c2+x2)2or,f′(x)=-4xc2(c2+x2)2

• MATH

f(x)=x2+c2x2−c2f(x)=x2+c2x2-c2 Find the derivative of the algebraic function, c is a…

f(x)=x2+c2x2−c2f(x)=x2+c2x2-c2 f'(x)=2x(x2−c2)−2x(x2+c2)(x2−c2)2f′(x)=2x(x2-c2)-2x(x2+c2)(x2-c2)2 or,f'(x)=−4xc2(x2−c2)2or,f′(x)=-4xc2(x2-c2)2

• MATH

f(x)=(x3−x)(x2+2)(x2+x−1)f(x)=(x3-x)(x2+2)(x2+x-1) Find the derivative of the algebraic function.

Since the function is a product, you need to find the derivative using the product rule, such that:

• MATH

f(x)=(2×3+5x)(x−3)(x+2)f(x)=(2×3+5x)(x-3)(x+2) Find the derivative of the algebraic function.

You need to use the product rule to evaluate the derivative of the function, such that: f'(x)=(2×3+5x)'(x−3)(x+2)+(2×3+5x)(x−3)'(x+2)+(2×3+5x)(x−3)(x+2)’f′(x)=(2×3+5x)′(x-3)(x+2)+(2×3+5x)(x-3)′(x+2)+(2×3+5x)(x-3)(x+2)′

• MATH

g(x)=(x2)((2x)−1x+1)g(x)=(x2)((2x)-1x+1) Find the derivative of the algebraic function.

You need to find derivative of the function using the product rule: f'(x)=((x2)’⋅(2x−1x+1)+(x2⋅(2x−1x+1)’f′(x)=((x2)′⋅(2x-1x+1)+(x2⋅(2x-1x+1)′ f'(x)=2x⋅(2x−1x+1)+(x2)⋅(−2×2+1(x+1)2)f′(x)=2x⋅(2x-1x+1)+(x2)⋅(-2×2+1(x+1)2)

• MATH

f(x)=2−(1x)x−3f(x)=2-(1x)x-3 Find the derivative of the algebraic function.

You need to find derivative of the function using the quotient rule: f'(x)=(2−1x)’⋅(x−3)−(2−1x)⋅(x−3)'(x−3)2f′(x)=(2-1x)′⋅(x-3)-(2-1x)⋅(x-3)′(x-3)2 f'(x)=x−3×2−2+1x(x−3)2f′(x)=x-3×2-2+1x(x-3)2

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h(x)=(x2+3)3h(x)=(x2+3)3 Find the derivative of the algebraic function.

Note:- 1) If y = x^n ; then dy/dx = y’ = n*x^(n-1) Now, h(x)=(x2+3)3h(x)=(x2+3)3h'(x)=3⋅2x⋅(x2+3)2h′(x)=3⋅2x⋅(x2+3)2 or,h'(x)=6x(x2+3)2or,h′(x)=6x(x2+3)2

• MATH

h(s)=(s3−2)2h(s)=(s3-2)2 Find the derivative of the algebraic function.

Note:- If y = x^n ; where n = constant ; then dy/dx = n*x^(n-1) Now, h(s)=(s3−2)2h(s)=(s3-2)2 h'(s)=(2)(3s2)⋅(s3−2)1h′(s)=(2)(3s2)⋅(s3-2)1 or,h'(s)=(6s2)⋅(s3−2)or,h′(s)=(6s2)⋅(s3-2) or,h'(s)=6s5−12s2or,h′(s)=6s5-12s2

• MATH

f(x)=(x−−√3)(x−−√+3)f(x)=(x3)(x+3) Find the derivative of the algebraic function.

f(x)=x13(x−−√+3)f(x)=x13(x+3) or,f(x)=x13(x12+3)or,f(x)=x13(x12+3) or,f(x)=x56+3x13or,f(x)=x56+3×13 f'(x)=(56)x−16+3⋅(13)x−23f′(x)=(56)x-16+3⋅(13)x-23 or,f'(x)=(56)⋅x−16+x−23or,f′(x)=(56)⋅x-16+x-23

• MATH

f(x)=3x−1x−−√f(x)=3x-1x Find the derivative of the algebraic function.

f(x)=3x−1x−−√f(x)=3x-1x or,f(x)=(3⋅x−−√)−(1x−−√)or,f(x)=(3⋅x)-(1x) or,f'(x)=(32)⋅(1x−−√)+(12)⋅(1×32)or,f′(x)=(32)⋅(1x)+(12)⋅(1×32) or,f'(x)={(12)⋅(1x−−√)⋅[3+(1x)]or,f′(x)={(12)⋅(1x)⋅[3+(1x)]

• MATH

f(x)=(x4)(1−2x+1)f(x)=(x4)(1-2x+1) Find the derivative of the algebraic function.

You need to find derivative of the function using the product rule: f'(x)=(x4)’⋅(x+1−2x+1)+(x4)⋅(x−1x+1)’f′(x)=(x4)′⋅(x+1-2x+1)+(x4)⋅(x-1x+1)′ You need to use the quotient rule to differentiate ((x-1)/(x+1)):

• MATH

f(x)=x(1−4x+3)f(x)=x(1-4x+3) Find the derivative of the algebraic function.

You need to evaluate the derivative of the function, hence you need to use the quotient rule, but first you need to find the common denominator within round parenthesis, such that:

• MATH

f(x)=x2+5x+6×2−4f(x)=x2+5x+6×2-4 Find the derivative of the algebraic function.

Given f(x)=x2+5x+6×2−4f(x)=x2+5x+6×2-4 Simplify the function to f(x)=(x+2)(x+3)(x−2)(x+2)=x+3x−2f(x)=(x+2)(x+3)(x-2)(x+2)=x+3x-2 Find the derivative using the Quotient Rule. f'(x)=(x−2)(1)−(x+3)(1)(x−2)2f′(x)=(x-2)(1)-(x+3)(1)(x-2)2…

• MATH

f(x)=4−3x−x2x2−1f(x)=4-3x-x2x2-1 Find the derivative of the algebraic function.

You need to find derivative of the function using the quotient rule: f'(x)=(4−3x−x2)’⋅(x2−1)−(4−3x−x2)⋅(x2−1)'(x2−1)2f′(x)=(4-3x-x2)′⋅(x2-1)-(4-3x-x2)⋅(x2-1)′(x2-1)2You need to use the quotient rule to differentiate…

• MATH

f(x)=sin(x)x,c=π6f(x)=sin(x)x,c=π6 Find f'(x) and f'(c).

Note:- 1) If y = sinx; then dy/dx = cosx 2) If y = x^n ; then dy/dx = n*x^(n-1) 3) If y = u/v ; where u & v are functions of ‘x’; then dy/dx = [vu’ – uv’]/(v^2) Now, f(x)=sinxxf(x)=sinxx

• MATH

f(x)=xcos(x),c=π4f(x)=xcos(x),c=π4 Find f'(x) and f'(c).

Note:- 1) If y = cosx; then dy/dx = -sinx 2) If y = x^n ; then dy/dx = n*x^(n-1) 3) If y = u*v ; where u & v are functions of ‘x’; then dy/dx = uv’ + vu’ Now, f(x)=xcos(x)f(x)=xcos(x)

• MATH

f(x)=x−4x+4,c=3f(x)=x-4x+4,c=3 Find f'(x) and f'(c).

f(x)=x−4x+4f(x)=x-4x+4 f'(x)=(x+4)−(x−4)(x+4)2f′(x)=(x+4)-(x-4)(x+4)2 or,f'(x)=8(x+4)2or,f′(x)=8(x+4)2 thus,f'(c)=f'(3)=8(3+4)2thus,f′(c)=f′(3)=8(3+4)2 or,f'(c)=f'(3)=872=849or,f′(c)=f′(3)=872=849

• MATH

f(x)=x2−4x−3,c=1f(x)=x2-4x-3,c=1 Find f'(x) and f'(c).

f(x)=x2−4x−3f(x)=x2-4x-3 f'(x)=(x−3)2x−(x2−4)(x−3)2f′(x)=(x-3)2x-(x2-4)(x-3)2 or,f'(x)=x2−6x+4(x−3)2or,f′(x)=x2-6x+4(x-3)2 Now,f'(c)=f'(1)=12−6⋅1+4(1−3)2Now,f′(c)=f′(1)=12-6⋅1+4(1-3)2 or,f'(c)=f'(1)=−14or,f′(c)=f′(1)=-14

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y=(x2−3x+2)(x3+1),c=2y=(x2-3x+2)(x3+1),c=2 Find f'(x) and f'(c).

You need to evaluate first y’, using the product rule, such that: y’=(x2−3x+2)'(x3+1)+(x2−3x+2)(x3+1)’y′=(x2-3x+2)′(x3+1)+(x2-3x+2)(x3+1)′ y’=(2x−3)(x3+1)+(x2−3x+2)(3×2)y′=(2x-3)(x3+1)+(x2-3x+2)(3×2)

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f(x)=(x3+4x)(3×2+2x−5),c=0f(x)=(x3+4x)(3×2+2x-5),c=0 Find f'(x) and f'(c).

You need to first find derivative of the function using the product rule: f'(x)=(x3+4x)’⋅(3×2+2x−5)+(x3+4x)⋅(3×2+2x−5)’f′(x)=(x3+4x)′⋅(3×2+2x-5)+(x3+4x)⋅(3×2+2x-5)′f'(x)=(3×2+4)⋅(3×2+2x−5)+(x3+4x)⋅(6x+2)f′(x)=(3×2+4)⋅(3×2+2x-5)+(x3+4x)⋅(6x+2)

• MATH

f(t)=cos(t)t3f(t)=cos(t)t3 Use the Quotient Rule to find the derivative of the function.

f(t)=cos(t)t3f(t)=cos(t)t3 f'(t)=(t3)⋅(−sin(t))−3⋅cos(t)⋅t2t6f′(t)=(t3)⋅(-sin(t))-3⋅cos(t)⋅t2t6 or,f'(t)=−(t3)sin(t)−3(t2)cos(t)t6or,f′(t)=-(t3)sin(t)-3(t2)cos(t)t6 or,f'(t)=−tsin(t)−3cos(t)t4or,f′(t)=-tsin(t)-3cos(t)t4