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mathematics
calculus early transcendentals 9th
Calculus Early Transcendentals 9th Edition James Stewart, Daniel K. Clegg, Saleem Watson, Lothar Redlin - Solutions
Compare the values of △y and dy if ϰ changes from 1 to 1.05. What if ϰ changes from 1 to 1.01? Does the approximation △y ≈ dy become better as △ϰ gets smaller?f (ϰ) = e2ϰ–2
Find an equation of the tangent line to the curve at the given point.y = ϰ + sin ϰ , (π, π)
Differentiate the function.k(r) = er + re
Find the derivative of the function.s(t) = √1 + sin t/ 1 + cos t
Use implicit differentiation to find an equation of the tangent line to the curve at the given point.tan(ϰ + y) + sec(ϰ – y) = 2, (π/8, π/8)
Differentiate.f (x) = x2ex/x2 + ex
Calculate y'.y = log5 (1 + 2x)
Compare the values of △y and dy if ϰ changes from 1 to 1.05. What if ϰ changes from 1 to 1.01? Does the approximation △y ≈ dy become better as △ϰ gets smaller?f (ϰ) = ϰ4 – ϰ + 1
Find an equation of the tangent line to the curve at the given point.y = sin ϰ + cos ϰ , (0, 1)
Show thatd/dϰ In(ϰ + √ϰ2 + 1) = 1/√ϰ2 + 1
Differentiate the function.j(ϰ) = ϰ2.4 + e2.4
Use implicit differentiation to find an equation of the tangent line to the curve at the given point.ye sin ϰ = ϰ cos y, (0, 0)
A particle moves along the curve y = 2 sin(πϰ/2). As the particle passes through the point (1/3, 1), its x-coordinate increases at a rate of √10 cm/s. How fast is the distance from the particle to the origin changing at this instant?
Differentiate.y = (z2 + ez) √z
Calculate y'. y = /sin /x
Compute △y and dy for the given values of ϰ and dϰ = △ϰ. Then sketch a diagram like Figure 5 showing the line segments with lengths dϰ, dy, and △y.y = eϰ, ϰ = 0, △ϰ = 0.5
Regard y as the independent variable and ϰ as the dependent variable and use implicit differentiation to find dϰ/dy.y sec ϰ = ϰ tan y
Differentiate the function.y = ln ϰa/bϰ
Calculate y'.sin(xy) = x2 − y
Compute △y and dy for the given values of ϰ and dϰ = △ϰ . Then sketch a diagram like Figure 5 showing the line segments with lengths dϰ , dy, and △y.y = √ϰ – 2 , ϰ = 3, △ϰ = 0.8
Show that d/dϰ (cot ϰ) = –csc2ϰ.
If cosh x = 5/3 and x > 0, find the values of the other hyperbolic functions at x.
Differentiate the function.G(r) = 3r3/2 + r5/2/r
Regard y as the independent variable and ϰ as the dependent variable and use implicit differentiation to find dϰ/dy.ϰ4y2 – ϰ3y + 2ϰy3 = 0
Two men stand 10 m apart on level ground near the edge of a cliff. One man drops a stone and one second later the other man drops a stone. One second after that, how fast is the distance between the two stones changing?Use the fact that the distance (in meters) a dropped stone falls after t seconds
Differentiate the function.y = ln√1 + 2ϰ/1 – 2ϰ
Differentiate.h(r) = aer/b + er
Some of the highest tides in the world occur in the Bay of Fundy on the Atlantic Coast of Canada. At Hopewell Cape the water depth at low tide is about 2.0 m and at high tide it is about 12.0 m. The natural period of oscillation is a little more than 12 hours and on a day in June, high tide
If tanh x = 12/13, find the values of the other hyperbolic functions at x.
Differentiate the function.y = √ϰ + ϰ /ϰ
Find the derivative of the function.y = (ϰ + 1/ϰ)5
If g(ϰ) + ϰ sin g(ϰ) = ϰ2, find g'(0).
Differentiate the function.h(ϰ) = eϰ2+ln ϰ
Use the fact that the distance (in meters) a dropped stone falls after t seconds is d = 4.9t2.A woman stands near the edge of a cliff and drops a stone over the edge. Exactly one second later she drops another stone. One second after that, how fast is the distance between the two stones changing?
Calculate y'.y = (1 − x-1)-1
Compute △y and dy for the given values of ϰ and dϰ = △ϰ. Then sketch a diagram like Figure 5 showing the line segments with lengths dϰ, dy, and △y.y = ϰ2 – 4ϰ, ϰ = 3, △ϰ = 0.5
Find the derivative of the function.y = √ϰ/ϰ + 1
Differentiate the function.f (x) = 3ϰ2 + ϰ3/ϰ
If f (ϰ) + ϰ2 [f(ϰ)]3 = 10 and f (1) = 2, find f' (1).
Differentiate the function.g(ϰ) = eϰ2 ln ϰ
(a) How long will it take an investment to double in value if the interest rate is 3%, compounded continuously?(b) What is the equivalent annual interest rate?
Differentiate.W(t) = et(1 + tet)
Calculate y'.y = sec(1 + x2)
Differentiate the function.S(R) = 4πR2
Find the derivative of the function.G(z) = (1 – 4z)2√z2 + 1
Find dy/dϰ by implicit differentiation.cos(ϰ2 + y2) = ϰey
Differentiate the function.y = ln(e–ϰ + ϰe–ϰ)
(a) If $4000 is invested at 1.75% interest, find the value of the investment at the end of 5 years if the interest is compounded(i) Annually, (ii) Semiannually, (iii) Monthly,(iv) Weekly, (v) Daily, (vi) Continuously.(b) If A(t) is the amount of the investment at time t for the
Differentiate.V(t) − (t + 2et) √t
Calculate y'.y = 3x ln x
Differentiate.f (θ) = θ cos θ sin θ
Differentiate the function.y = 3eϰ + 4/3√ϰ
Find the derivative of the function.F(ϰ) = (4ϰ + 5)3(ϰ2 – 2x + 5)4
(a) If $2500 is borrowed at 4.5% interest, find the amounts due at the end of 3 years if the interest is compounded(i) Annually, (ii) Quarterly, (iii) Monthly,(iv) Weekly,(v) Daily, (vi) Hourly, (vii) Continuously.(b) Suppose $2500 is borrowed and the interest is compounded
Calculate y'.y = ex sec x
Differentiate.g(z) = z/sec z + tan z
If a cylindrical water tank holds 5000 gallons, and the water drains from the bottom of the tank in 40 minutes, then Torricelli’s Law gives the volume V of water remaining in the tank after t minutes asV = 5000(1 – 1/40 t)2 0 ≤ t ≤ 40Find the rate at which water is draining from the tank
Differentiate the function.F(t) = (2t – 3)2
Find the derivative of the function.A(r) = √r · er2+1
Differentiate the function.y = ln |3 – 2ϰ5|
Differentiate.H(u) = (u − √u )(u + √u )
Calculate y'. t y = tan 1 + t?
Differentiate.y = t sin t/1 + t
Differentiate the function.f (x) = x3(x + 3)
Find dy/dϰ by implicit differentiation.√ϰ + y = ϰ4 + y4
Find dy/dϰ by implicit differentiation.sin ϰ cos y = ϰ2 – 5y
Differentiate the function.g(t) = ln t(t2 + 1)4/3√ 2t – 1
Find the differential of the function.y = eϰ/1 – eϰ
Differentiate the function.W(t) = √t − 2et
Differentiateh(w) = (w2 + 3w)(w–1 – w–4)
Find the derivative of the function.y = ϰ2e–3ϰ
Find dy/dϰ by implicit differentiation.2ϰey + yeϰ = 3
Calculate y'.y = √arctan x
Differentiate.f (w) = 1 + sec w/1 – sec w
Differentiate the function.t(ϰ) = 1/√ϰ + 4√ϰ
Find the derivative of the function.y = 5√ϰ
Differentiate.y = √ϰ/√ϰ + 1
Calculate y'. 4 и — 1 u y = и? + и + 1
Find the differential of the function.y = √1 + cos θ
Differentiate the function.h(w) = √2 w – √2
Find the derivative of the function.g(ϰ) = eϰ2–ϰ
Differentiate.y = s – √s/s2
Find the differential of the function.y = 1/ϰ2 – 3ϰ
Determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example that disproves the statement.If g(x) = x5, then g(x) – g(2) lim x2 80. x - 2
Differentiate.y = ϰ/2 – tan ϰ
Prove the identity.sinh(x + y) = sinh x cosh y + cosh x sinh y
Differentiate the function.y = 2ϰ + √ϰ
Find the derivative of the function.g(θ) = cos2 θ
Find dy/dϰ by implicit differentiation.tan(ϰ – y) = 2ϰy3 + 1
Differentiate.F(ϰ) = 1/2ϰ3 – 6ϰ2 + 5
Differentiate the function.y = log10 sec ϰ
(a) What quantities are given in the problem?(b) What is the unknown?(c) Draw a picture of the situation for any time t.(d) Write an equation that relates the quantities.(e) Finish solving the problem.If a snowball melts so that its surface area decreases at a rate of 1 cm2/min, find the rate at
Determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example that disproves the statement.An equation of the tangent line to the parabola y = x2 at (–2, 4) is y – 4 = 2x(x + 2).
Ifcalculate f(46)(3). Express your answer using factorial notation:n! = 1 · 2 · 3 · ∙ ∙ ∙ · (n – 1) · n x46 + x45 + 2 f(x) 1 + x
Differentiate.y = cos ϰ/1 – sin ϰ
Prove the identity.cosh x − sinh x = e-x
Differentiate the function.r(t) = a/t2 + b/t4
Find the derivative of the function.f (θ) = cos(θ2)
Differentiate.f (t) = 5t/t3 – t – 1
Differentiate the function.y = log8(ϰ2 + 3ϰ)
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