|BSM

Questions

1 / 10

Time

Score

Find The Derivative Of A Trigonometric Function Using The Chain Rule

Use chain rule to find f’(x), if f(x)= cos cos (x²)

2xcos(x²)

2cos(x²)

xcos(x²)

Use chain rule to find f’(x), if $f\left(x\right)=\sin\sqrt{x}$

$\frac{1}{2}\cos\sqrt{x}$

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

$\frac{1}{2}\sqrt{x}\cos\sqrt{x}$

Use chain rule to find 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 f’(x), if f(X)= sinsin (e^x2)

Cosx²e^x3 cos(e^x3)

cos3x² e^x2 cos(e^x2)

cos2xe^x2 cos (e^x2)

Use chain rule to find f’(x), if f(x)=cos cos (tanx)

2xcos(x²)

sinsin(tanx)x

-sinsin(tanx)x

Use chain rule to find dy/dx, if y=sin2x

2cos2x

2 sinx

cos2x

If f(x) and g(x) are differentiable then $\frac{d}{dx}\left[fog\left(x\right)\right]$ =

F’[g(x)]

f’[g(x)]g’(x)

f’[g(x)]g(x)

Use chain rule to find f’(x), if f(x) = sin sin(e^x3)

Cosx²e^x3 cos(e^x3)

cos 3x² e^x3cos(e^x3)

-(4x-6)cos cos (2x²-6x)

Use chain rule to find f’(x), if f(X) = 2 tan tan (x²)

2x sec(x²)

2sec(x²)

4xsec²(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 Trigonometric Function Using The Chain Rule

Use chain rule to find f’(x), if f(x)= cos cos (x2)

2xcos(x2)

2cos(x2)

xcos(x2)

Use chain rule to find f’(x), if ﻿f(x)=sin⁡xf\left(x\right)=\sin\sqrt{x}f(x)=sinx​﻿

﻿12cos⁡x\frac{1}{2}\cos\sqrt{x}21​cosx​﻿

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

﻿12xcos⁡x\frac{1}{2}\sqrt{x}\cos\sqrt{x}21​x​cosx​﻿

Use chain rule to find 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 f’(x), if f(X)= sinsin (ex2)

Cosx2ex3 cos(ex3)

cos3x2 ex2 cos(ex2)

cos2xex2 cos (ex2)

Use chain rule to find f’(x), if f(x)=cos cos (tanx)

2xcos(x2)

sinsin(tanx)x

-sinsin(tanx)x

Use chain rule to find dy/dx, if y=sin2x

2cos2x

2 sinx

cos2x

If f(x) and g(x) are differentiable then ﻿ddx[fog(x)]\frac{d}{dx}\left[fog\left(x\right)\right]dxd​[fog(x)]﻿ =

F’[g(x)]

f’[g(x)]g’(x)

f’[g(x)]g(x)

Use chain rule to find f’(x), if f(x) = sin sin(ex3)

Cosx2ex3 cos(ex3)

cos 3x2 ex3 cos( ex3)

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

Use chain rule to find f’(x), if f(X) = 2 tan tan (x2)

2x sec(x2)

2sec(x2)

4xsec2(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!

Use chain rule to find f’(x), if f(x)= cos cos (x²)

2xcos(x²)

2cos(x²)

xcos(x²)

Use chain rule to find f’(x), if $f\left(x\right)=\sin\sqrt{x}$

$\frac{1}{2}\cos\sqrt{x}$

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

$\frac{1}{2}\sqrt{x}\cos\sqrt{x}$

Use chain rule to find 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 f’(x), if f(X)= sinsin (e^x2)

Cosx²e^x3 cos(e^x3)

cos3x² e^x2 cos(e^x2)

cos2xe^x2 cos (e^x2)

2xcos(x²)

If f(x) and g(x) are differentiable then $\frac{d}{dx}\left[fog\left(x\right)\right]$ =

Use chain rule to find f’(x), if f(x) = sin sin(e^x3)

Cosx²e^x3 cos(e^x3)

cos 3x² e^x3cos(e^x3)

-(4x-6)cos cos (2x²-6x)

Use chain rule to find f’(x), if f(X) = 2 tan tan (x²)

2x sec(x²)

2sec(x²)

4xsec²(x²)

$\frac{dy}{du}$

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

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