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Derivadas trigonométricas y derivación implícita12B

Authored by Jhon Jaramillo

Mathematics

12th Grade

Used 2+ times

Derivadas trigonométricas y derivación implícita12B
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6 questions

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1.

MULTIPLE CHOICE QUESTION

15 mins • 1 pt

La derivada de la función

 y=(2x2)(cosx2)y=\left(2-x^2\right)\left(\cos x^2\right)  

 y=2x[cosx2+(2x2)senx2]y'=-2x\left[\cos x^2+\left(2-x^2\right)senx^2\right]  

 y=2x[cosx2+(2x2)senx2]y'=2x\left[\cos x^2+\left(2-x^2\right)senx^2\right]  

 y=2x[cosx2(2x2)senx2]y'=-2x\left[\cos x^2-\left(2-x^2\right)senx^2\right]  

 y=2x[cosx2+(2+x2)senx2]y'=-2x\left[\cos x^2+\left(2+x^2\right)senx^2\right]  

2.

MULTIPLE CHOICE QUESTION

15 mins • 1 pt

Al derivar la función

 x2+2xyy2=1x^2+2xy-y^2=1  obtenemos

 y=x+yxyy'=\frac{-x+y}{x-y}  

 y=xyxyy'=\frac{-x-y}{x-y}  

 y=xyx+yy'=\frac{-x-y}{x+y}  

 y=x+yxyy'=\frac{x+y}{x-y}  

3.

MULTIPLE CHOICE QUESTION

15 mins • 1 pt

La ecuación de la recta tangente a la curva en el punto (3,3)

 y3+x3=6xyy^3+x^3=6xy  es

 y=x+6y=x+6  

 y=x6y=x-6  

 y=x+6y=-x+6  

 y=x6y=-x-6  

4.

MULTIPLE CHOICE QUESTION

15 mins • 1 pt

La derivada de la función

 y=senxtanxy=senx\tan x  en términos de seno y coseno es

 y=senx(1+cos2x)cos2xy'=\frac{senx\left(1+\cos^2x\right)}{\cos^2x}  

 y=senx(1cos2x)cos2xy'=\frac{senx\left(1-\cos^2x\right)}{\cos^2x}  

 y=senx(1+cos2x)cos2xy'=\frac{-senx\left(1+\cos^2x\right)}{\cos^2x}  

 y=senx(1cos2x)cos2xy'=\frac{-senx\left(1-\cos^2x\right)}{\cos^2x}  

5.

MULTIPLE CHOICE QUESTION

15 mins • 1 pt

Al derivar la función

 y=sen3(4x2+x)y=sen^3\left(4x^2+x\right)  es

 y=(24x+3)[sen2(4x2+x)]cos(4x2)y'=\left(24x+3\right)\left[sen^2\left(4x^2+x\right)\right]\cos\left(4x^2\right)  

 y=(24x+3)[sen2(4x2+x)]cos(4x2x)y'=\left(24x+3\right)\left[sen^2\left(4x^2+x\right)\right]\cos\left(4x^2-x\right)  

 y=(24x+3)[sen2(4x2+x)]cos(4x2+x)y'=\left(24x+3\right)\left[sen^2\left(4x^2+x\right)\right]\cos\left(4x^2+x\right)  

 y=(24x)[sen2(4x2+x)]cos(4x2+x)y'=\left(24x\right)\left[sen^2\left(4x^2+x\right)\right]\cos\left(4x^2+x\right)  

6.

MULTIPLE CHOICE QUESTION

15 mins • 1 pt

Al derivar  la función

 xy=cos(x+y)xy=\cos\left(x+y\right)  obtenemos

 y=y+sen(xy)x+sen(x+y)y'=-\frac{y+sen\left(x-y\right)}{x+sen\left(x+y\right)}  

 y=ysen(x+y)xsen(x+y)y'=-\frac{y-sen\left(x+y\right)}{x-sen\left(x+y\right)}  

 y=y+sen(x+y)x+sen(x+y)y'=\frac{y+sen\left(x+y\right)}{x+sen\left(x+y\right)}  

 y=y+sen(x+y)x+sen(x+y)y'=-\frac{y+sen\left(x+y\right)}{x+sen\left(x+y\right)}  

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