Carilah persamaan garis singgung kurva y = e^x/(1+ x^2) di titik (1, 1/2e) !

Posted by andi telaumbanua on Feb 11, 2018 in Matematika |
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Carilah persamaan garis singgung kurva y = e^x/((1+ x^2)) di titik (1, 1/2e) !

Jawab:

Cari gradien kurva m = y_((1,1/2 e))^’

Maka :

dy/dx = (((d (e^x ))/dx) (1+ x^2 )-(d(1+ x^2 )/dx )(e^x))/(1+ x^2)^2

dy/dx = ((e^x )(1+ x^2 )- (2x)(e^x))/(1+ x^2)^2

dy/dx = (e^x (1+ x^2- 2x ))/(1+ x^2)^2

Maka :

m = y_((1,1/2 e))^’

m = (e^x (1+ x^2- 2x ))/(1+ x^2 )^2 dititik (1,1/2 e)

m = (e^1 (1+ 1^2- 2(1) ))/(1+ 1^2 )^2

m = 0

maka persamaan garis singgungnya adalah :

y -y_(1 ) = m ( x – x_1)

y – 1/2 e=0(x-1)

y – 1/2 e=0

y = 1/2 e

karena m = 0

maka: persamaan garis singgung kurva di titik (1,1/2 e) berupa garis horizontal dan konstan yaitu:

y = 1/2 e

for more detailed writing click on the following link Carilah persamaan garis singgung kurva y = e^x/((1+ x^2)) di titik (1, 1/2e) !

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