braindrill

The Chain Rule

10 practice questionsAP Calculus ABStep-by-step solutions

Try a few

1

Differentiate y=sin3(ln(cosx))y = \sin^3(\ln(\cos x)) with respect to xx.

A.3sec(x)sin2(ln(cosx))cos(ln(cosx))3\sec(x)\sin^2(\ln(\cos x))\cos(\ln(\cos x))
B.3tan(x)sin2(ln(cosx))cos(ln(cosx))3\tan(x)\sin^2(\ln(\cos x))\cos(\ln(\cos x))
C.3tan(x)sin2(ln(cosx))cos(ln(cosx))-3\tan(x)\sin^2(\ln(\cos x))\cos(\ln(\cos x))
D.3tan(x)sin2(ln(cosx))-3\tan(x)\sin^2(\ln(\cos x))
🔒 Answer + full step-by-step solutionUnlock free →
2

A particle moves along a line so that its position at time tt (in seconds) is given by s(t)=cos(t2+1)s(t) = \cos\left(\sqrt{t^2 + 1}\right) meters. Find the velocity function v(t)=s(t)v(t) = s'(t).

A.sin(t2+1)2t2+1-\frac{\sin\left(\sqrt{t^2+1}\right)}{2\sqrt{t^2+1}} m/s
B.2tsin(t2+1)-2t\sin\left(\sqrt{t^2+1}\right) m/s
C.tsin(t2+1)t2+1-\frac{t\sin\left(\sqrt{t^2+1}\right)}{\sqrt{t^2+1}} m/s
D.tsin(t2+1)t2+1\frac{t\sin\left(\sqrt{t^2+1}\right)}{\sqrt{t^2+1}} m/s
🔒 Answer + full step-by-step solutionUnlock free →
3

A spherical balloon is being inflated. Its radius at time tt seconds is r(t)=2t+1r(t)=\sqrt{2t+1} cm. The volume of a sphere is V=43πr3V=\frac{4}{3}\pi r^3. Determine the rate of change of volume with respect to time, dVdt\frac{dV}{dt}, as an expression in tt.

A.2π2t+12\pi\sqrt{2t+1} cm3^3/s
B.4π(2t+1)3/24\pi(2t+1)^{3/2} cm3^3/s
C.4π2t+14\pi\sqrt{2t+1} cm3^3/s
D.8π2t+18\pi\sqrt{2t+1} cm3^3/s
🔒 Answer + full step-by-step solutionUnlock free →

Master the chain rulenot just preview it

All 10 questions with worked solutions, an AI tutor that explains every step, and games that make the drilling stick. Free to start.

Practice this topic free

No card needed · 10 free AI questions daily