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Questions

1 / 10

Time

Score

Find The Derivative Of A Composite Function Using The Chain Rule

Use chain rule to evaluate d/dx [f(x)], if f(x)= cos cos(tanx)

2xcos(x²)

sinsin(tanx)x

-sinsin(tanx)x

Use chain rule to evaluate d/dx [f(x)] if $f\left(x\right)=\sqrt{x+\sqrt{x}}$

$\frac{2\sqrt{x}}{4\sqrt{x+\sqrt{x}\ \sqrt{x}}}$

$\frac{\left(2\sqrt{x}+1\right)}{4\sqrt{x+\sqrt{x}\ \sqrt{x}}}$

$\frac{\left(\sqrt{x}+1\right)}{4\sqrt{x+\sqrt{x}\ \sqrt{x}}}$

Use chain rule to evaluate d/dx [f(x)], if f(x)=sinsin(2x²-6x)

$\left(4x-6\right)\cos\ \cos\left(2x^2-6x\right)$

$\cos\cos\left(2x^2-6x\right)$

$-\left(4x-6\right)\cos\cos\left(2x^2-6x\right)$

Use chain rule to evaluate d/dx [f(x)], if $f\left(x\right)=\sqrt{\tan\left(1+x^2\right)}$

$\frac{\sec^2\left(x^2+1\right)}{\sqrt{\tan\left(1+x^2\right)}}$

$\ \frac{x\sec^2\left(x^2+1\right)}{\sqrt{\tan\left(1+x^2\right)}}$

$\ \frac{x\sec^2\left(x^2+1\right)}{\sqrt{\tan\left(x^2\right)}}$

Use chain rule to find dy/dx. if y=f(u)=u+1and x=u+(1/u)

$\frac{2u^2}{u^2-1}$

$\frac{u^2}{u^2+1}$

$\frac{u^2}{u^2-1}$

Use chain rule to evaluate d/dx [f(x)], if f(x)=sinsin(e^x3)

Cos x²e^x3 cos(e^x3)

cos3x²e^x3 cos(e^x3)

-(4x-6) cos cos (2x² -6x)

Use chain rule to evaluate d/dx [f(x)], if f(x)= tantan(x²)

2x sec(x²)

2sec(x²)

2x sec²(x²)

Use chain rule to evaluate d/dx [f(x)], if f(x)=sinsin(e^x2)

Cosx² e^x3 cos(e^x3)

cos3x² e^x2 cos(e^x2)

cos 2xe^x2 cos(e^x2)

Use chain rule to evaluate d/dx [f(x)], if f(x)= sin sin(x²)

2xcos(x²)

2cos(x²)

xcos(x²)

If y=f(u) and u=g(x). Calculate dy/dx

$\frac{dy}{du}$

$\frac{dy}{dx}\ \frac{dx}{du}$

$\frac{dy}{du}\ \frac{du}{dx}$

Questions

Time

Score

Find The Derivative Of A Composite Function Using The Chain Rule

Use chain rule to evaluate d/dx [f(x)], if f(x)= cos cos(tanx)

2xcos(x2)

sinsin(tanx)x

-sinsin(tanx)x

Use chain rule to evaluate d/dx [f(x)] if ﻿f(x)=x+xf\left(x\right)=\sqrt{x+\sqrt{x}}f(x)=x+x​​﻿

﻿2x4x+x x\frac{2\sqrt{x}}{4\sqrt{x+\sqrt{x}\ \sqrt{x}}}4x+x​ x​​2x​​﻿

﻿(2x+1)4x+x x\frac{\left(2\sqrt{x}+1\right)}{4\sqrt{x+\sqrt{x}\ \sqrt{x}}}4x+x​ x​​(2x​+1)​﻿

﻿(x+1)4x+x x\frac{\left(\sqrt{x}+1\right)}{4\sqrt{x+\sqrt{x}\ \sqrt{x}}}4x+x​ x​​(x​+1)​﻿

Use chain rule to evaluate d/dx [f(x)], if f(x)=sinsin(2x2-6x)

﻿(4x−6)cos⁡ cos⁡(2x2−6x)\left(4x-6\right)\cos\ \cos\left(2x^2-6x\right)(4x−6)cos cos(2x2−6x)﻿

﻿cos⁡cos⁡(2x2−6x)\cos\cos\left(2x^2-6x\right)coscos(2x2−6x)﻿

﻿−(4x−6)cos⁡cos⁡(2x2−6x)-\left(4x-6\right)\cos\cos\left(2x^2-6x\right)−(4x−6)coscos(2x2−6x)﻿

Use chain rule to evaluate d/dx [f(x)], if ﻿f(x)=tan⁡(1+x2)f\left(x\right)=\sqrt{\tan\left(1+x^2\right)}f(x)=tan(1+x2)​﻿

﻿sec⁡2(x2+1)tan⁡(1+x2)\frac{\sec^2\left(x^2+1\right)}{\sqrt{\tan\left(1+x^2\right)}}tan(1+x2)​sec2(x2+1)​﻿

﻿ xsec⁡2(x2+1)tan⁡(1+x2)\ \frac{x\sec^2\left(x^2+1\right)}{\sqrt{\tan\left(1+x^2\right)}} tan(1+x2)​xsec2(x2+1)​﻿

﻿ xsec⁡2(x2+1)tan⁡(x2)\ \frac{x\sec^2\left(x^2+1\right)}{\sqrt{\tan\left(x^2\right)}} tan(x2)​xsec2(x2+1)​﻿

Use chain rule to find dy/dx. if y=f(u)=u+1and x=u+(1/u)

﻿2u2u2−1\frac{2u^2}{u^2-1}u2−12u2​﻿

﻿u2u2+1\frac{u^2}{u^2+1}u2+1u2​﻿

﻿u2u2−1\frac{u^2}{u^2-1}u2−1u2​﻿

Use chain rule to evaluate d/dx [f(x)], if f(x)=sinsin(ex3)

Cos x2ex3 cos(ex3)

cos3x2ex3 cos(ex3)

-(4x-6) cos cos (2x2 -6x)

Use chain rule to evaluate d/dx [f(x)], if f(x)= tantan(x2)

2x sec(x2)

2sec(x2)

2x sec2(x2)

Use chain rule to evaluate d/dx [f(x)], if f(x)=sinsin(ex2)

Cosx2 ex3 cos(ex3)

cos3x2 ex2 cos(ex2)

cos 2xex2 cos(ex2)

Use chain rule to evaluate d/dx [f(x)], if f(x)= sin sin(x2)

2xcos(x2)

2cos(x2)

xcos(x2)

If y=f(u) and u=g(x). Calculate dy/dx

﻿dydu\frac{dy}{du}dudy​﻿

﻿dydx dxdu\frac{dy}{dx}\ \frac{dx}{du}dxdy​ dudx​﻿

﻿dydu dudx\frac{dy}{du}\ \frac{du}{dx}dudy​ dxdu​﻿

Incorrect

Great!

2xcos(x²)

Use chain rule to evaluate d/dx [f(x)] if $f\left(x\right)=\sqrt{x+\sqrt{x}}$

$\frac{2\sqrt{x}}{4\sqrt{x+\sqrt{x}\ \sqrt{x}}}$

$\frac{\left(2\sqrt{x}+1\right)}{4\sqrt{x+\sqrt{x}\ \sqrt{x}}}$

$\frac{\left(\sqrt{x}+1\right)}{4\sqrt{x+\sqrt{x}\ \sqrt{x}}}$

Use chain rule to evaluate d/dx [f(x)], if f(x)=sinsin(2x²-6x)

$\left(4x-6\right)\cos\ \cos\left(2x^2-6x\right)$

$\cos\cos\left(2x^2-6x\right)$

$-\left(4x-6\right)\cos\cos\left(2x^2-6x\right)$

Use chain rule to evaluate d/dx [f(x)], if $f\left(x\right)=\sqrt{\tan\left(1+x^2\right)}$

$\frac{\sec^2\left(x^2+1\right)}{\sqrt{\tan\left(1+x^2\right)}}$

$\ \frac{x\sec^2\left(x^2+1\right)}{\sqrt{\tan\left(1+x^2\right)}}$

$\ \frac{x\sec^2\left(x^2+1\right)}{\sqrt{\tan\left(x^2\right)}}$

$\frac{2u^2}{u^2-1}$

$\frac{u^2}{u^2+1}$

$\frac{u^2}{u^2-1}$

Use chain rule to evaluate d/dx [f(x)], if f(x)=sinsin(e^x3)

Cos x²e^x3 cos(e^x3)

cos3x²e^x3 cos(e^x3)

-(4x-6) cos cos (2x² -6x)

Use chain rule to evaluate d/dx [f(x)], if f(x)= tantan(x²)

2x sec(x²)

2sec(x²)

2x sec²(x²)

Use chain rule to evaluate d/dx [f(x)], if f(x)=sinsin(e^x2)

Cosx² e^x3 cos(e^x3)

cos3x² e^x2 cos(e^x2)

cos 2xe^x2 cos(e^x2)

Use chain rule to evaluate d/dx [f(x)], if f(x)= sin sin(x²)

2xcos(x²)

2cos(x²)

xcos(x²)

$\frac{dy}{du}$

$\frac{dy}{dx}\ \frac{dx}{du}$

$\frac{dy}{du}\ \frac{du}{dx}$