f(t)=5−t−−−−√f(t)=5-t Find the derivative of the function.

# f(t)=5−t−−−−√f(t)=5-t Find the derivative of the function.

• MATH

g(θ)=sec((12)θ)tan((12)θ)g(θ)=sec((12)θ)tan((12)θ) Find the derivative of the function.

Given the function g(θ)=sec(θ2)tan(θ2)g(θ)=sec(θ2)tan(θ2) . We have to find its derivative. Here we will apply the product rule i.e., ddθ(uv)=uv’+u’vddθ(uv)=uv′+u′v So here we have,

• MATH

h(x)=sin(2x)cos(2x)h(x)=sin(2x)cos(2x) Find the derivative of the function.

Note:- 1) 2sin(ax)*cos(ax) = sin2(ax) 2) If y = sinx ; then dy/dx = cosx Now, h(x)=sin(2x)⋅cos(2x)h(x)=sin(2x)⋅cos(2x) or,h(x)=(12)⋅sin(4x)or,h(x)=(12)⋅sin(4x) thus,h'(x)=(42)cos(4x)thus,h′(x)=(42)cos(4x) or,h'(x)=2cos(4x)or,h′(x)=2cos(4x)

• MATH

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

You need to evaluate the derivative of the function, hence you need to use the chain rule, differentiating first the cosine, then the power function and then the argument, such that:

• MATH

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

You need to use the chain rule to evaluate the derivative of the function, such that: y’=(sin(π⋅x)2)’⋅((π⋅x)2)’⋅(π⋅x)’y′=(sin(π⋅x)2)′⋅((π⋅x)2)′⋅(π⋅x)′ y’=(cos(π⋅x)2)⋅(2(π⋅x))⋅πy′=(cos(π⋅x)2)⋅(2(π⋅x))⋅π y’=2π2⋅x(cos(π⋅x)2)y′=2π2⋅x(cos(π⋅x)2)Hence,…

• MATH

h(x)=sec(x2)h(x)=sec(x2) Find the derivative of the function.

You need to use the chain rule to evaluate the derivative of the function, such that: h'(x)=(sec(x)2)’⋅((x)2)’h′(x)=(sec(x)2)′⋅((x)2)′ h'(x)=2x(sec(x)2)⋅(tan(x)2)h′(x)=2x(sec(x)2)⋅(tan(x)2) h'(x)=2x(sec(x)2)⋅(tan(x)2)h′(x)=2x(sec(x)2)⋅(tan(x)2) Hence,…

• MATH

g(x)=5tan(3x)g(x)=5tan(3x) Find the derivative of the function.

You need to use the chain rule to evaluate the derivative of the function, such that: g'(x)=(5tan(3x))’⋅(3x)’g′(x)=(5tan(3x))′⋅(3x)′ g'(x)=(5cos2(3x))⋅3g′(x)=(5cos2(3x))⋅3 g'(x)=(15cos2(3x))g′(x)=(15cos2(3x)) Hence, evaluating the…

• MATH

y=sin(πx)y=sin(πx) Find the derivative of the function.

Given: y=sin(πx)y=sin(πx) To find the derivative of the function use the chain rule. y’=πcos(πx)y′=πcos(πx)

• MATH

y=cos(4x)y=cos(4x) Find the derivative of the function.

Given: y=cos(4x)y=cos(4x) To find the derivative of the function use the chain rule. y’=−4sin(4x)y′=-4sin(4x)

• MATH

g(x)=(2+(x2+1)4)3g(x)=(2+(x2+1)4)3 Find the derivative of the function.

g(x)=(2+(x2+1)4)3g(x)=(2+(x2+1)4)3 g'(x)=[3(2+(x2+1)4)2]⋅[4(x2+1)3]⋅[2x]g′(x)=[3(2+(x2+1)4)2]⋅[4(x2+1)3]⋅[2x] or,g'(x)=24x(x2+1)3[2+(x2+1)4]2or,g′(x)=24x(x2+1)3[2+(x2+1)4]2

• MATH

f(x)=((x2+3)5+x)2f(x)=((x2+3)5+x)2 Find the derivative of the function.

You need to find derivative of the function using the chain rule: f'(x)=(((x2+3)5+x)2)’⋅((x2+3)5+x)’f′(x)=(((x2+3)5+x)2)′⋅((x2+3)5+x)′ f'(x)=(2((x2+3)5+x))⋅(5(x2+3)4⋅(x2+3)’+1)f′(x)=(2((x2+3)5+x))⋅(5(x2+3)4⋅(x2+3)′+1)

• MATH

g(x)=(3×2−22x+3)3g(x)=(3×2-22x+3)3 Find the derivative of the function.

Given g(x)=(3×2−22x+3)3g(x)=(3×2-22x+3)3 Find the derivative using the Quotient Rule within the Chain Rule. g'(x)=3(3×2−22x+3)2[(2x+3)(6x)−(3×2−2)(2)(2x+3)4]g′(x)=3(3×2-22x+3)2[(2x+3)(6x)-(3×2-2)(2)(2x+3)4]…

• MATH

f(v)=(1−2v1+v)3f(v)=(1-2v1+v)3 Find the derivative of the function.

Given f(v)=(1−2v1+v)3f(v)=(1-2v1+v)3 Find the derivative using the Quotient Rule within the Chain Rule. f'(v)=3(1−2v1+v)2[(1+v)(−2)−(1−2v)(1)(1+v)2]f′(v)=3(1-2v1+v)2[(1+v)(-2)-(1-2v)(1)(1+v)2] f'(v)=3(1−2v)2(−2−2v−1+2v)(1+v)4f′(v)=3(1-2v)2(-2-2v-1+2v)(1+v)4…

• MATH

h(t)=(t2t3+2)2h(t)=(t2t3+2)2 Find the derivative of the function.

Given: h(t)=(t2t3+2)2h(t)=(t2t3+2)2 Find the derivative of the function by using the Quotient Rule within the Chain Rule. h'(t)=2[t2t3+2][(t3+2)(2t)−(t2)(3t2)(t3+2)2]h′(t)=2[t2t3+2][(t3+2)(2t)-(t2)(3t2)(t3+2)2]…

• MATH

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

Given: g(x)=(x+5×2+2)2g(x)=(x+5×2+2)2 Find the derivative by using the Quotient Rule within the Chain Rule. g'(x)=2[x+5×2+2][(x2+2)(1)−(x+5)(2x)(x2+2)2]g′(x)=2[x+5×2+2][(x2+2)(1)-(x+5)(2x)(x2+2)2]…

• MATH

y=xx4+4−−−−−√y=xx4+4 Find the derivative of the function.

You need to find derivative of the function using quotient rule: y’=x’⋅(x4+4−−−−−√)−x⋅(x4+4−−−−−√)’x4+4y′=x′⋅(x4+4)-x⋅(x4+4)′x4+4 y’=1⋅(x4+4−−−−−√)−x⋅(x4+4)’2×4+4√x4+4y′=1⋅(x4+4)-x⋅(x4+4)′2×4+4×4+4

• MATH

y=xx2+1−−−−−√y=xx2+1 Find the derivative of the function.

Given y=xx2+1−−−−−√y=xx2+1 Rewrite the function as y=x(x2+1)−12y=x(x2+1)-12 Find the derivative using the Product Rule and Chain Rule. y’=x[(−12)(x2+1)−32]+(x2+1)−12(1)y′=x[(-12)(x2+1)-32]+(x2+1)-12(1)…

• MATH

y=(12)(x2)16−x2−−−−−−√y=(12)(x2)16-x2 Find the derivative of the function.

You need to use the product and chain rules to evaluate the derivative of the function, such that: f'(x)=(12)(x2)'(16−x2−−−−−−√)+(12)(x2)(16−x2−−−−−−√)’f′(x)=(12)(x2)′(16-x2)+(12)(x2)(16-x2)′

• MATH

y=x1−x2−−−−−√y=x1-x2 Find the derivative of the function.

You need to use the product and chain rules to evaluate the derivative of the function, such that: f'(x)=(x)'(1−x2−−−−−√)+(x)(1−x2−−−−−√)’f′(x)=(x)′(1-x2)+(x)(1-x2)′

• MATH

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

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

• MATH

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

You need to use the product and chain rules to evaluate the derivative of the function, such that: f'(x)=(x2)'(x−2)4+(x2)((x−2)4)’f′(x)=(x2)′(x-2)4+(x2)((x-2)4)′ f'(x)=(2x)(x−2)4+4(x2)((x−2)3)⋅(x−2)’f′(x)=(2x)(x-2)4+4(x2)((x-2)3)⋅(x-2)′…

• MATH

g(t)=1t2−2−−−−−√g(t)=1t2-2 Find the derivative of the function.

g(t)=1t2−2−−−−−√g(t)=1t2-2  or,g(t)=(t2−2)−12or,g(t)=(t2-2)-12 Thus,g'(t)=−(12)⋅(2t)⋅(t2−2)−32Thus,g′(t)=-(12)⋅(2t)⋅(t2-2)-32 or,g'(t)=−t⋅(t2−2)−32or,g′(t)=-t⋅(t2-2)-32 or,g'(t)=−t(t2−2)32or,g′(t)=-t(t2-2)32

• MATH

y=13x+5−−−−−−√y=13x+5 Find the derivative of the function.

y=13x+5−−−−−−√=(3x+5)−12y=13x+5=(3x+5)-12 thus,dydx=y’=−(12)⋅{(3x+5)−32}⋅3thus,dydx=y′=-(12)⋅{(3x+5)-32}⋅3 or,dydx=−(32)⋅(3x+5)−32or,dydx=-(32)⋅(3x+5)-32 or,dydx=y’=−(32)⋅{1(3x+5)32}or,dydx=y′=-(32)⋅{1(3x+5)32}

• MATH

y=−(3(t−2)4)y=-(3(t-2)4) Find the derivative of the function.

Given: y=−(3(t−2)4)y=-(3(t-2)4) Rewrite the equation as y=−3(t−2)−4y=-3(t-2)-4 Find the derivative using the Chain Rule. y’=12(t−2)−5y′=12(t-2)-5 y’=12(t−2)5y′=12(t-2)5

• MATH

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

f(t)=1(t−3)2f(t)=1(t-3)2 or,f(t)=(t−3)−2or,f(t)=(t-3)-2 Thus,f'(t)=−2(t−3)−3Thus,f′(t)=-2(t-3)-3 or,f'(t)=−2(t−3)3or,f′(t)=-2(t-3)3

• MATH

s(t)=14−5t−t2s(t)=14-5t-t2 Find the derivative of the function.

s(t)=14−5t−t2s(t)=14-5t-t2 or,s'(t)=−−5−2t[4−5t−t2]2or,s′(t)=–5-2t[4-5t-t2]2 or,s'(t)=5+2t[4−5t−t2]2or,s′(t)=5+2t[4-5t-t2]2

• MATH

y=1x−2y=1x-2 Find the derivative of the function.

Given: y=1x−2y=1x-2 Rewrite the function as y=1(x−2)−1y=1(x-2)-1 Find the derivative using the Chain Rule. y’=−1(x−2)−2y′=-1(x-2)-2 y’=−1(x−2)2y′=-1(x-2)2

• MATH

f(x)=12x−5−−−−−−√3f(x)=12x-53 Find the derivative of the function.

Given: f(x)=12x−5−−−−−−√3f(x)=12x-53 Rewrite the function as f(x)=(12x−5)13f(x)=(12x-5)13 Find the derivative by using the Chain Rule. f'(x)=(13)(12x−5)−23(12)f′(x)=(13)(12x-5)-23(12) f'(x)=4(12x−5)−23f′(x)=4(12x-5)-23…

• MATH

y=29−x2−−−−−√4y=29-x24 Find the derivative of the function.

Given y=29−x2−−−−−√4y=29-x24 Rewrite the function as y=2(9−x2)14y=2(9-x2)14 Find the derivative using the Chain Rule. y’=(14)(2)(9−x2)−34(−2x)y′=(14)(2)(9-x2)-34(-2x) y’=−1x(9−x2)−34y′=-1x(9-x2)-34 y’=−x(9−x2)34y′=-x(9-x2)34…

• MATH

f(x)=x2−4x+2−−−−−−−−−−√f(x)=x2-4x+2 Find the derivative of the function.

Given f(x)=x2−4x+2−−−−−−−−−−√f(x)=x2-4x+2 Rewrite the function as f(x)=(x2−4x+2)12f(x)=(x2-4x+2)12 Find the derivative using the Chain Rule. f'(x)=(12)(x2−4x+2)−12(2x−4)f′(x)=(12)(x2-4x+2)-12(2x-4) f'(x)=(12)2(x−2)(x2−4x+2)12f′(x)=(12)2(x-2)(x2-4x+2)12…

• MATH

y=6×2+1−−−−−−√3y=6×2+13 Find the derivative of the function.

Given y=6×2+1−−−−−−√3y=6×2+13 Rewrite the function as y=(6×2+1)13y=(6×2+1)13 Find the derivative using the Chain Rule. y’=(13)(6×2+1)−23(12x)y′=(13)(6×2+1)-23(12x) y’=4x(6×2+1)−23y′=4x(6×2+1)-23 y’=4x(6×2+1)23y′=4x(6×2+1)23…

• MATH

g(x)=4−3×2−−−−−−√g(x)=4-3×2 Find the derivative of the function.

• MATH

f(t)=5−t−−−−√f(t)=5-t Find the derivative of the function.

f(t)=5−t−−−−√f(t)=5-t f'(t)=125−t−−−−√⋅(−1)f′(t)=125-t⋅(-1) or,f'(t)=−125−t−−−−√or,f′(t)=-125-t   Note:- If y = sqrt(x) ; then dy/dx = 1/{2sqrt(x)}

• MATH

f(t)=(9t+2)23f(t)=(9t+2)23 Find the derivative of the function.

f(t)=(9t+2)23f(t)=(9t+2)23 f'(t)=(23)⋅9⋅(9t+2)−13f′(t)=(23)⋅9⋅(9t+2)-13 or,f'(t)=6⋅(9t+2)−13or,f′(t)=6⋅(9t+2)-13

• MATH

g(x)=3(4−9x)4g(x)=3(4-9x)4 Find the derivative of the function.

Given: g(x)=3(4−9x)4g(x)=3(4-9x)4 Find the derivative of the function by using the Chain Rule. g'(x)=12(4−9x)3(−9)g′(x)=12(4-9x)3(-9) g'(x)=−108(4−9x)3g′(x)=-108(4-9x)3

• MATH

y=5(2−x3)4y=5(2-x3)4 Find the derivative of the function.

Given: y=5(2−x3)4y=5(2-x3)4 To find the derivative of the function use the Chain Rule. y’=20(2−x3)3(−3×2)y′=20(2-x3)3(-3×2) y’=−60×2(2−x3)3y′=-60×2(2-x3)3

• MATH

y=(4x−1)3y=(4x-1)3 Find the derivative of the function.

Given: y=(4x−1)3y=(4x-1)3 To find the derivative of the function use the Chain Rule. y’=3(4x−1)2(4)y′=3(4x-1)2(4) y’=12(4x−1)2y′=12(4x-1)2

• MATH

f4(x)=2x+1,f6(x)f4(x)=2x+1,f6(x) Find the given higher-order derivative.

Given:- f””(x)=2x+1f′′′′(x)=2x+1 Thus,f””'(x)=2Thus,f′′′′′(x)=2 andf”””(x)=0andf′′′′′′(x)=0

• MATH

f”'(x)=2x−−√,f4(x)f′′′(x)=2x,f4(x) Find the given higher-order derivative.

Given f”'(x)=2x−−√f′′′(x)=2x Thus,f””(x)=22x−−√Thus,f′′′′(x)=22x or,f””(x)=1x−−√or,f′′′′(x)=1x

• MATH

f”(x)=2−(2x),f”'(x)f′′(x)=2-(2x),f′′′(x) Find the given higher-order derivative.

Given: f”(x)=2−(2x)f′′(x)=2-(2x) Rewrite the function as f”(x)=2−2x−1f′′(x)=2-2x-1 f”'(x)=2x−2f′′′(x)=2x-2 f”'(x)=2x2f′′′(x)=2×2

• MATH

f'(x)=x2,f”(x)f′(x)=x2,f′′(x) Find the given higher-order derivative.

Given: f'(x)=x2f′(x)=x2 f”(x)=2xf′′(x)=2x

• MATH

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

f(x)=sec(x)f(x)=sec(x) f'(x)=sec(x)tan(x)f′(x)=sec(x)tan(x) f”(x)=sec(x)⋅sec2(x)+tan(x)⋅sec(x)tan(x)f′′(x)=sec(x)⋅sec2(x)+tan(x)⋅sec(x)tan(x) or,f”(x)=sec(x)[sec2(x)+tan2(x)]or,f′′(x)=sec(x)[sec2(x)+tan2(x)]

• MATH

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

f(x)=xsin(x)f(x)=xsin(x) f'(x)=xcos(x)+sin(x)f′(x)=xcos(x)+sin(x) f”(x)=−xsin(x)+cos(x)+cos(x)f′′(x)=-xsin(x)+cos(x)+cos(x) or,f”(x)=2cos(x)−xsin(x)or,f′′(x)=2cos(x)-xsin(x)

• MATH

f(x)=x2+3xx−4f(x)=x2+3xx-4 Find the second derivative of the function.

f(x)=x2+3xx−4f(x)=x2+3xx-4 f'(x)=(x−4)(2x+3)−(x2+3x)(x−4)2f′(x)=(x-4)(2x+3)-(x2+3x)(x-4)2 or,f'(x)=2×2+3x−8x−12−x2−3x(x−4)2or,f′(x)=2×2+3x-8x-12-x2-3x(x-4)2 or,f'(x)=x2−8x−12(x−4)2or,f′(x)=x2-8x-12(x-4)2 Now,Now,

• MATH

f(x)=xx−1f(x)=xx-1 Find the second derivative of the function.

f(x)=xx−1f(x)=xx-1 f'(x)=(x−1)−x(x−1)2f′(x)=(x-1)-x(x-1)2 or,f'(x)=−1(x−1)2or,f′(x)=-1(x-1)2f”(x)=2(x−1)3f′′(x)=2(x-1)3 or,f”(x)=2(x−1)−3or,f′′(x)=2(x-1)-3

• MATH

f(x)=x2+3x−3f(x)=x2+3x-3 Find the second derivative of the function.

You need to evaluate the first derivative of the function, using the quotient rule for 3x−3:3x-3: f'(x)=2x−9x2x6f′(x)=2x-9x2x6 You need to simplify by x2:x2: f'(x)=2x−9x4f′(x)=2x-9×4 You need to…

• MATH

f(x)=4x32f(x)=4×32 Find the second derivative of the function.

Note:- If x = x^n ; where n = constant ; then dy/dx = n*x^(n-1) Now, f(x)=4x32f(x)=4×32 f'(x)=4⋅(32)⋅x12f′(x)=4⋅(32)⋅x12 or,f'(x)=6⋅x12or,f′(x)=6⋅x12 f”(x)=6⋅(12)⋅x−12f′′(x)=6⋅(12)⋅x-12 or,f”(x)=3⋅x−12or,f′′(x)=3⋅x-12

• MATH

f(x)=4×5−2×3+5x2f(x)=4×5-2×3+5×2 Find the second derivative of the function.

You need to evaluate the first derivative of the function: f'(x)=(4×5−2×3+5×2)’f′(x)=(4×5-2×3+5×2)′ f'(x)=(4×5)’−(2×3)’+(5×2)’f′(x)=(4×5)′-(2×3)′+(5×2)′ f'(x)=20×4−6×2+10xf′(x)=20×4-6×2+10xYou need to evaluate the second derivative of…

• MATH

f(x)=x4+2×3−3×2−xf(x)=x4+2×3-3×2-x Find the second derivative of the function.

Note:- If y = x^n ; where n = constant ; then dy/dx = n*x^(n-1) If y = k ; where k = constant ; then dy/dx = 0 Now, f(x)=x4+2×3−3×2−xf(x)=x4+2×3-3×2-x f'(x)=4×3+6×2−6x−1f′(x)=4×3+6×2-6x-1

• MATH

f(x)=x−4×2−7f(x)=x-4×2-7 Determine the point(s) at which the graph of the function has a…

Given: f(x)=x−4×2−7f(x)=x-4×2-7 Find the derivative of the function using the Quotient Rule. Set the derivative equal to zero and solve for the critical x value(s). When the derivative is zero the…