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

# h(t)=2(cot(πt+2))2h(t)=2(cot(πt+2))2 Find the derivative of the function.

• MATH

4cos(x)sin(y)=14cos(x)sin(y)=1 Find dydxdydx by implicit differentiation.

4cos(x)⋅sin(y)=14cos(x)⋅sin(y)=1 Differentiating both sides w.r.t ‘x’ we get −4sin(x)sin(y)+4cos(x)cos(y)(dydx)=0-4sin(x)sin(y)+4cos(x)cos(y)(dydx)=0 or,dydx=sin(x)sin(y)cos(x)cos(y)or,dydx=sin(x)sin(y)cos(x)cos(y) or,dydx=y’=tan(x)⋅tan(y)or,dydx=y′=tan(x)⋅tan(y)

• MATH

x3−3(x2)y+2x(y2)=12×3-3(x2)y+2x(y2)=12 Find dydxdydx by implicit differentiation.

x3−3x2y+2xy2=12×3-3x2y+2xy2=12 Differentiating both sides with respect to x, 3×2−3(x2dydx+yddx(x2))+2(x(2y)dydx+y2)=03×2-3(x2dydx+yddx(x2))+2(x(2y)dydx+y2)=0 3×2−3(x2dydx+y(2x))+4xydydx+2y2=03×2-3(x2dydx+y(2x))+4xydydx+2y2=0 3×2−3x2dydx−6xy+4xydydx+2y2=03×2-3x2dydx-6xy+4xydydx+2y2=0…

• MATH

xy−−√=(x2)y+1xy=(x2)y+1 Find dydxdydx by implicit differentiation.

xy−−√=x2y+1xy=x2y+1 Differentiating both sides with respect to x, 12(xy)−12ddx(xy)=x2ddx(y)+yddx(x2)12(xy)-12ddx(xy)=x2ddx(y)+yddx(x2) 12(xy−−√)(xdydx+y)=x2dydx+y(2x)12(xy)(xdydx+y)=x2dydx+y(2x)…

• MATH

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

(x3)(y3)−y=x(x3)(y3)-y=x Differentiating both sides with respect to x, ddx(x3y3−y)=ddx(x)ddx(x3y3-y)=ddx(x) x3ddx(y3)+y3ddx(x3)−ddx(y)=1x3ddx(y3)+y3ddx(x3)-ddx(y)=1 x3(3y2dydx)+y3(3×2)−dydx=1×3(3y2dydx)+y3(3×2)-dydx=1 3x3y2dydx+3x2y3−dydx=13x3y2dydx+3x2y3-dydx=1…

• MATH

(x2)y+(y2)x=−2(x2)y+(y2)x=-2 Find dydxdydx by implicit differentiation.

x2y+y2x=−2x2y+y2x=-2 Differentiating both sides with respect to x, x2ddx(y)+yddx(x2)+y2ddx(x)+xddx(y2)=0x2ddx(y)+yddx(x2)+y2ddx(x)+xddx(y2)=0 x2dydx+y(2x)+y2+x(2y)dydx=0x2dydx+y(2x)+y2+x(2y)dydx=0 x2dydx+2xydydx+2xy+y2=0x2dydx+2xydydx+2xy+y2=0…

• MATH

x3−xy+y2=7×3-xy+y2=7 Find dydxdydx by implicit differentiation.

Given: x3−xy+y2=7×3-xy+y2=7 3×2−[xdydx+y(1)]+2ydydx=03×2-[xdydx+y(1)]+2ydydx=0 3×2−xdydx−y+2ydydx=03×2-xdydx-y+2ydydx=0 −xdydx+2ydydx=−3×2+y-xdydx+2ydydx=-3×2+y xdydx−2ydydx=3×2−yxdydx-2ydydx=3×2-y dydx(x+2y)=3×2−ydydx(x+2y)=3×2-y dydx=3×2−yx+2ydydx=3×2-yx+2y

• MATH

2×3+3y3=642×3+3y3=64 Find dydxdydx by implicit differentiation.

2×3+3y3=642×3+3y3=64  Differentiating both sides w.r.t ‘x’ we get 2⋅3⋅x2+3⋅3⋅y2⋅(dydx)=02⋅3⋅x2+3⋅3⋅y2⋅(dydx)=0 or,2×2+3y2(dydx)=0or,2×2+3y2(dydx)=0 or,dydx=−(23)⋅(xy)2or,dydx=-(23)⋅(xy)2

• MATH

x12+y12=16×12+y12=16 Find dydxdydx by implicit differentiation.

x12+y12=16×12+y12=16 or,(12)x−12+(12)y−12(dydx)=0or,(12)x-12+(12)y-12(dydx)=0 or,dydx=−(xy)−12=−(yx)12or,dydx=-(xy)-12=-(yx)12

• MATH

x2−y2=25×2-y2=25 Find dydxdydx by implicit differentiation.

Given: x2−y2=25×2-y2=25 2x−2ydydx=02x-2ydydx=0 −2ydydx=−2x-2ydydx=-2x dydx=−2x−2ydydx=-2x-2y dydx=xydydx=xy

• MATH

x2+y2=9×2+y2=9 Find dydxdydx by implicit differentiation.

Given: x2+y2=9×2+y2=9 2x+2ydydx=02x+2ydydx=0 2ydydx=−2x2ydydx=-2x dydx=−2x2ydydx=-2x2y dydx=−xydydx=-xy

• MATH

g(t)=tan(2t),((π6),3–√)g(t)=tan(2t),((π6),3) Evaluate the second derivative of the function at the…

g(t)=tan(2t)g(t)=tan(2t) differentiating with respect to t, g'(t)=sec2(2t)⋅2g′(t)=sec2(2t)⋅2 differentiating again, g”(t)=2(2sec(2t))(sec(2t)tan(2t)⋅2)g′′(t)=2(2sec(2t))(sec(2t)tan(2t)⋅2) g”(t)=8tan(2t)sec2(2t)g′′(t)=8tan(2t)sec2(2t) Now to evaluate the g”(t)…

• MATH

f(x)=cos(x2),(0,1)f(x)=cos(x2),(0,1) Evaluate the second derivative of the function at the given point….

We will need to use chain rule (f(g(x)))’=f'(g(x))⋅g'(x)(f(g(x)))′=f′(g(x))⋅g′(x) First we find the first derivative. f'(x)=−sin(x2)⋅2x=−2xsin(x2)f′(x)=-sin(x2)⋅2x=-2xsin(x2) Now we calculate second derivative, but for that we will…

• MATH

f(x)=1x+4−−−−−√,(0,(12))f(x)=1x+4,(0,(12)) Evaluate the second derivative of the function at the…

You need to evaluate the second derivative of the function, hence, you need first to evaluate the first derivative, using the quotient and the chain rules, such that:

• MATH

h(x)=(19)(3x+1)3,(1,(649))h(x)=(19)(3x+1)3,(1,(649)) Evaluate the second derivative of the function at the…

You need first to evaluate the first derivative of the function, using chain rule, such that: h'(x)=((19)(3x+1)3)’h′(x)=((19)(3x+1)3)′ h'(x)=((39)(3x+1)2)⋅(3x+1)’h′(x)=((39)(3x+1)2)⋅(3x+1)′ h'(x)=((13)(3x+1)2)⋅(3)h′(x)=((13)(3x+1)2)⋅(3)…

• MATH

f(x)=(sec(πx))2f(x)=(sec(πx))2 Find the second derivative of the function.

f(x)=(sec(πx))2=sec2(πx)f(x)=(sec(πx))2=sec2(πx) f'(x)=2πsec(πx)⋅tan(πx)⋅sec(πx)=2πsec2(πx)⋅tan(πx)f′(x)=2πsec(πx)⋅tan(πx)⋅sec(πx)=2πsec2(πx)⋅tan(πx)f”(x)=2π2⋅sec4(πx)+4π2sec2(πx)⋅tan2(πx)f′′(x)=2π2⋅sec4(πx)+4π2sec2(πx)⋅tan2(πx)

• MATH

f(x)=sin(x2)f(x)=sin(x2) Find the second derivative of the 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) Now, f(x)=sin(x2)f(x)=sin(x2) f'(x)=2x⋅cos(x2)f′(x)=2x⋅cos(x2) f”(x)=2⋅cos(x2)−4(x2)⋅sin(x2)f′′(x)=2⋅cos(x2)-4(x2)⋅sin(x2)

• MATH

f(x)=8(x−2)2f(x)=8(x-2)2 Find the second derivative of the function.

Note:- If y = (ax+b)^n ; where a,b,n are constants ; then dy/dx = na*x^(n-1) Now, y=8(x−2)2y=8(x-2)2 or,y=8⋅(x−2)−2or,y=8⋅(x-2)-2 or,y=−16(x−2)−3or,y=-16(x-2)-3 or,y=48(x−2)−4or,y=48(x-2)-4 or,y=48(x−2)4or,y=48(x-2)4

• MATH

f(x)=1x−6f(x)=1x-6 Find the second derivative of the function.

Note:- if y = (ax+b)^n ; where a,b,n are constants ; then dy/dx = an*(ax+b)^(n-1) Now, f(x)=1x−6=(x−6)−1f(x)=1x-6=(x-6)-1 or,f'(x)=−1(x−6)−2or,f′(x)=-1(x-6)-2 or,f”(x)=2(x−6)−3or,f′′(x)=2(x-6)-3 or,f”(x)=2(x−6)3or,f′′(x)=2(x-6)3

• MATH

f(x)=6(x3+4)3f(x)=6(x3+4)3 Find the second derivative of the function.

Note:- If y = x^n ; where n = constant ; then dy/dx = n*x^(n-1) Now, f(x)=6(x3+4)3f(x)=6(x3+4)3 thus,f'(x)=6⋅3⋅3(x2)⋅(x3+4)2thus,f′(x)=6⋅3⋅3(x2)⋅(x3+4)2 or,f'(x)=54(x2)(x3+4)2or,f′(x)=54(x2)(x3+4)2 Differentiating again w.r.t ‘x’ we…

• MATH

f(x)=5(2−7x)4f(x)=5(2-7x)4 Find the second derivative of the function.

f(x)=5(2−7x)4f(x)=5(2-7x)4 f'(x)=5⋅4⋅(−7)⋅(2−7x)3f′(x)=5⋅4⋅(-7)⋅(2-7x)3 or,f'(x)=−140(2−7x)3or,f′(x)=-140(2-7x)3 Thus,f”(x)=−140⋅3⋅(−7)⋅(2−7x)2Thus,f′′(x)=-140⋅3⋅(-7)⋅(2-7x)2 or,f”(x)=2940⋅(2−7x)2or,f′′(x)=2940⋅(2-7x)2

• MATH

y=2(tan(x))3,((π4),2)y=2(tan(x))3,((π4),2) Find an equation of the tangent line to the graph of f at the given…

You need to find the equation of the tangent line to the given curve, at the point (pi/4,2), using the formula: f(x)−f(π4)=f'(π4)(x−π4)f(x)-f(π4)=f′(π4)(x-π4) You need to notice that f(π4)=2.f(π4)=2. You need…

• MATH

f(x)=(tan(x))2,((π4),1)f(x)=(tan(x))2,((π4),1)Find an equation of the tangent line to the graph of f at the…

You need to find the equation of the tangent line to the given curve, at the point (π4,1)(π4,1) , using the formula: f(x)−f(π4)=f'(π4)(x−π4)f(x)-f(π4)=f′(π4)(x-π4) You need to notice that f(π4)=1.f(π4)=1. You…

• MATH

y=cos(3x),((π4),−2–√2)y=cos(3x),((π4),-22) (a) Find an equation of the tangent line to the graph of f at…

The function y = cos (3x). The slope of the tangent to the graph of this function at (π4,−2–√2)(π4,-22) is y+2√2x−π4=y'(π4)y+22x-π4=y′(π4) y’=−3⋅sin(3⋅x)y′=-3⋅sin(3⋅x) y'(π4)=−32–√y′(π4)=-32 The…

• MATH

f(x)=sin(2x),(π,0)f(x)=sin(2x),(π,0) (a) Find an equation of the tangent line to the graph of f at the…

The function f(x) = sin(2x). The slope of a line tangent to this curve at the point (pi,0) is given by the equation y−0x−π=f'(π)y-0x-π=f′(π) f'(x)=(sin(2x))’f′(x)=(sin(2x))′ = 2*cos(2x) f'(π)=2f′(π)=2…

• MATH

f(x)=(9−x2)23,(1,4)f(x)=(9-x2)23,(1,4) (a) Find an equation of the tangent line to the graph of f…

The function f(x)=(9−x2)23f(x)=(9-x2)23 . At the point (1, 4) the equation of the tangent to this curve is: y−4x−1=f'(1)y-4x-1=f′(1) f(x)=(9−x2)23f(x)=(9-x2)23

• MATH

y=(4×3+3)2,(−1,1)y=(4×3+3)2,(-1,1) (a) Find an equation of the tangent line to the graph of f at the…

You need to find the equation of the tangent line to the given curve, at the point (-1,1), using the formula: f(x)−f(−1)=f'(−1)(−(−1))f(x)-f(-1)=f′(-1)(-(-1)) You need to put y = f(x) and you need to notice that…

• MATH

f(x)=(13)xx2+5−−−−−√,(2,2)f(x)=(13)xx2+5,(2,2) (a) Find an equation of the tangent line to the graph of…

The function f(x)=(13)⋅x⋅x2+5−−−−−√f(x)=(13)⋅x⋅x2+5 . The slope of a line tangent to this curve at x = a is f'(a). The derivative f'(x)=(13)⋅(x⋅x2+5−−−−−√)’f′(x)=(13)⋅(x⋅x2+5)′ =

• MATH

f(x)=2×2−7−−−−−−√,(4,5)f(x)=2×2-7,(4,5) (a) Find an equation of the tangent line to the graph of f at…

f(x)=2×2−7−−−−−−√f(x)=2×2-7 Slope of the tangent line is the derivative of the function at that point. f'(x)=(12)(2×2−7)−12(4x)f′(x)=(12)(2×2-7)-12(4x) f'(x)=2x2x2−7−−−−−−√f′(x)=2x2x2-7 Slope of the tangent line (m) at (4,5)…

• MATH

y=(1x)+cos(x)−−−−−√,(π22π)y=(1x)+cos(x),(π22π) Find and evaluate the derivative of the function at…

Hello! y(x)=(1x)+cos(x)−−−−−√y(x)=(1x)+cos(x) . cos(x) must be >=0 and cos(x) < 0 for π>x>π2π>x>π2 (at the right neighborhood of π2π2 ). So there can be only the left derivative. Check that…

• MATH

y=26−(sec(4x))3,(0,25)y=26-(sec(4x))3,(0,25) Find and evaluate the derivative of the function at the given…

You need to evaluate the derivative of the function, using the chain rule, such that: y’=(26−(sec(4x))3)’⇒y’=−3(sec(4x))2⋅(sec(4x))⋅(tan(4x))⋅(4x)’y′=(26-(sec(4x))3)′⇒y′=-3(sec(4x))2⋅(sec(4x))⋅(tan(4x))⋅(4x)′y’=−12(sec(4x))3⋅(tan(4x))y′=-12(sec(4x))3⋅(tan(4x))…

• MATH

f(x)=x+42x−5,(9,1)f(x)=x+42x-5,(9,1) Find and evaluate the derivative of the function at the…

You need to evaluate the equation of the tangent line at (9,1), using the formula: f(x) – f(9) = f'(9)(x – 1) Notice that f(9) = 1. You need to evaluate f'(x), using the quotient rule, such that:…

• MATH

f(t)=3t+2t−1,(0,−2)f(t)=3t+2t-1,(0,-2) Find and evaluate the derivative of the function at the…

First we differentiate the function using the following rule (fg)’=f’g−fg’g2(fg)′=f′g-fg′g2 f'(t)=3(t−1)−(3t+2)⋅1(t−1)2=3t−3−3t−2(t−1)2=−5(t−1)2f′(t)=3(t-1)-(3t+2)⋅1(t-1)2=3t-3-3t-2(t-1)2=-5(t-1)2Now we evaluate the derivative at 0….

• MATH

f(x)=1(x2−3x)2,(4,(116))f(x)=1(x2-3x)2,(4,(116)) Find and evaluate the derivative of the function at the…

You need to evaluate the derivative of the function, using the chain and quotient rules, such that: y’=−2(x2−3x)(x2−3x)'(x2−3x)4⇒y’=−2(x2−3x)(2x−3)(x2−3x)4y′=-2(x2-3x)(x2-3x)′(x2-3x)4⇒y′=-2(x2-3x)(2x-3)(x2-3x)4…

• MATH

f(x)=5×3−2,(−2,(−12))f(x)=5×3-2,(-2,(-12)) Find and evaluate the derivative of the function at the…

The function f(x) = 5/(x^3 – 2) = 5*(x^3 – 2)^-1. The derivative f'(x) = 5*(x^3 – 2)^-2*-1*3x^2 = (-15x^2)/(x^3 – 2)^2 At x = -2, f'(x) = (-15*4)/(-8 – 2)^2 = -60/100 = -0.6 The equation of the…

• MATH

y=3×3+4x−−−−−−−√5,(2,2)y=3×3+4×5,(2,2) Find and evaluate the derivative of the function at the given…

You need to evaluate the derivative of the function, using the chain rule, such that: y’=((3×3+4x)15)’⇒y’=((15)(3×3+4x)15−1)(3×3+4x)’y′=((3×3+4x)15)′⇒y′=((15)(3×3+4x)15-1)(3×3+4x)′

• MATH

y=x2+8x−−−−−−√,(1,3)y=x2+8x,(1,3) Find and evaluate the derivative of the function at the given…

You need to evaluate the derivative of the function, using the chain rule, such that: y’=((x2+8x)12)’⇒y’=((12)(x2+8x)12−1)(x2+8x)’y′=((x2+8x)12)′⇒y′=((12)(x2+8x)12-1)(x2+8x)′

• MATH

y=cos(sin(tan(πx))−−−−−−−−−−√)y=cos(sin(tan(πx))) Find the derivative of the function.

Given the function y=cos(sin(tan(πx))−−−−−−−−−−√)y=cos(sin(tan(πx))) . We have to find the derivative. Let us begin, dydx=−sin(sin(tan(πx))−−−−−−−−−−√).ddx[sin(tan(πx))−−−−−−−−−−√]−−−−−→(1)dydx=-sin(sin(tan(πx))).ddx[sin(tan(πx))]—–→(1)Now,…

• MATH

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

y=sin(tan(2x))y=sin(tan(2x)) dydx=[dsin(tan(2x))dx]⋅[dtan(2x)dx]⋅[d2xdx]dydx=[dsin(tan(2x))dx]⋅[dtan(2x)dx]⋅[d2xdx] or,dydx=y’=[cos(tan(2x))]⋅[sec2(2x)]⋅2or,dydx=y′=[cos(tan(2x))]⋅[sec2(2x)]⋅2 or,dydx=y’=2sec2(2x)⋅cos(tan(2x))or,dydx=y′=2sec2(2x)⋅cos(tan(2x))

• MATH

y=sin(x−−√3)+sin(x)−−−−−√3y=sin(x3)+sin(x)3 Find the derivative of the function.

Hello! We have the composite functions, for such a functions [f(g(x))]’=f'(g(x))⋅g'(x).[f(g(x))]′=f′(g(x))⋅g′(x). Also recall that [sin(y)]’=cos(y)[sin(y)]′=cos(y)and [y13]’=(13)⋅y−23.[y13]′=(13)⋅y-23. Let’s start:

• MATH

y=x−−√+(14)(sin(2x)2)y=x+(14)(sin(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, where it is required, such that: y’=(x−−√)’+(14)⋅(sin(2x)2)’⋅((2x)2)’⋅(2x)’y′=(x)′+(14)⋅(sin(2x)2)′⋅((2x)2)′⋅(2x)′

• MATH

y=3x−5cos((πx)2)y=3x-5cos((πx)2) Find the derivative of the function.

y=3x−5cos((πx)2)y=3x-5cos((πx)2) dydx=y’=3+5[sin((πx)2)]⋅[2πx]⋅πdydx=y′=3+5[sin((πx)2)]⋅[2πx]⋅π or,dydx=y’=3+10(π2)⋅x⋅sin((πx)2)or,dydx=y′=3+10(π2)⋅x⋅sin((πx)2)

• MATH

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

f(t)=3(sec(πt−1))2f(t)=3(sec(πt-1))2 or,f(t)=3sec2(πt−1)or,f(t)=3sec2(πt-1) thus,f'(t)=6sec(πt−1)⋅[sec(πt−1)⋅tan(πt−1)]⋅πthus,f′(t)=6sec(πt-1)⋅[sec(πt-1)⋅tan(πt-1)]⋅π or,f'(t)=6π⋅sec2(πt−1)⋅tan(πt−1)or,f′(t)=6π⋅sec2(πt-1)⋅tan(πt-1)

• MATH

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

h(t)=2(cot(πt+2))2=2cot2(πt+2)h(t)=2(cot(πt+2))2=2cot2(πt+2) h'(t)=4cot(πt+2)⋅[−cosec2(πt+2)]⋅πh′(t)=4cot(πt+2)⋅[-cosec2(πt+2)]⋅π or,h'(t)=−4π⋅cot(πt+2)⋅cosec2(πt+2)or,h′(t)=-4π⋅cot(πt+2)⋅cosec2(πt+2)

• MATH

f(θ)=(14)(sin(2θ))2f(θ)=(14)(sin(2θ))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 power, then the sine function and then the argument, such that:

• MATH

g(θ)=(cos(8θ))2g(θ)=(cos(8θ))2 Find the derivative of the function.

g(θ)={cos(8θ)}2=cos2(8θ)g(θ)={cos(8θ)}2=cos2(8θ) g'(θ)=2⋅cos(8θ)⋅(−sin(8θ))⋅8g′(θ)=2⋅cos(8θ)⋅(-sin(8θ))⋅8 or,g'(θ)=−16sin(8θ)cos(8θ)or,g′(θ)=-16sin(8θ)cos(8θ) or,g'(θ)=−8⋅sin(16θ)or,g′(θ)=-8⋅sin(16θ)

• MATH

f(θ)=(tan(5θ))2f(θ)=(tan(5θ))2 Find the derivative of the function.

f(θ)={tan5(θ)}2=tan2(5θ)f(θ)={tan5(θ)}2=tan2(5θ) thus,f'(θ)=2⋅tan(5θ)⋅sec2(5θ)⋅5thus,f′(θ)=2⋅tan(5θ)⋅sec2(5θ)⋅5 or,f'(θ)=10⋅tan(5θ)⋅sec2(5θ)or,f′(θ)=10⋅tan(5θ)⋅sec2(5θ)

• MATH

g(t)=5(cos(πt))2g(t)=5(cos(πt))2 Find the derivative of the function.

g(t) = 5(cos(πt))25(cos(πt))2 g'(t)=5⋅2⋅cos(πt)⋅{−sin(πt)}⋅πg′(t)=5⋅2⋅cos(πt)⋅{-sin(πt)}⋅π or,g'(t)=−10cos(πt)⋅sin(πt)or,g′(t)=-10cos(πt)⋅sin(πt) or,g'(t)=−5sin(2πt)or,g′(t)=-5sin(2πt)Note:- 1)2sint⋅cost=sin2t2sint⋅cost=sin2t

• MATH

y=4(sec(x))2y=4(sec(x))2 Find the derivative of the function.

y=4sec2(x)y=4sec2(x) dydx=4⋅2⋅sec(x)⋅sec(x)⋅tan(x)dydx=4⋅2⋅sec(x)⋅sec(x)⋅tan(x) or,dydx=8sec2(x)tan(x)or,dydx=8sec2(x)tan(x) Note:- 1) If y = sec(x) ; then dy/dx = secx*tanx

• MATH

g(v)=cos(v)csc(v)g(v)=cos(v)csc(v) Find the derivative of the function.

Given: g(v)=cos(v)csc(v)g(v)=cos(v)csc(v) The original function equivalent to g(v)=cos(v)sin(v).g(v)=cos(v)sin(v). g'(x)=cos(v)cos(v)+sin(v)(−sin(v))g′(x)=cos(v)cos(v)+sin(v)(-sin(v)) g'(x)=cos2(v)−sin2(v)g′(x)=cos2(v)-sin2(v)