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junior college 1 | H2 Maths
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Hello! How do i solve this qns? Sorry for all the differentiation qns, im having a bit of a struggle haha
Date Posted: 4 years ago
Views: 277
tan-¹x + tan-¹y + tan-¹xy = 7π/12
when x = 1,
tan-¹ 1 + tan-¹y + tan-¹y = 7π/12
π/4 + 2 tan-¹y = 7π/12
2 tan-¹y = π/3
tan-¹y = π/6
y = tan π/6
= 1/√3
= √3 / 3
when x = 1,
tan-¹ 1 + tan-¹y + tan-¹y = 7π/12
π/4 + 2 tan-¹y = 7π/12
2 tan-¹y = π/3
tan-¹y = π/6
y = tan π/6
= 1/√3
= √3 / 3
tan-¹x + tan-¹y + tan-¹xy = 7π/12
Differentiating with respect to x,
1/(1 + x²) + 1/(1 + y²) dy/dx + d/dx (tan-¹xy) = 0
d/dx (tan-¹xy) = - 1/(1 + x²) - 1/(1 + y²)dy/dx
Differentiating with respect to x,
1/(1 + x²) + 1/(1 + y²) dy/dx + d/dx (tan-¹xy) = 0
d/dx (tan-¹xy) = - 1/(1 + x²) - 1/(1 + y²)dy/dx
d/dx (tan-¹xy) = - 1/(1 + x²) - 1/(1 + y²)dy/dx
1/(1 + (xy)²) (y + x dy/dx) = - 1/(1 + x²) - 1/(1 + y²)dy/dx
When x = 1 , y = √3/3
(recall part i's working)
1/(1 + (√3/3)²) (√3/3 + dy/dx) = -1/(1 + 1²) - 1(1 + (√3/3)²) dy/dx
¾(√3/3 + dy/dx)= -1/2 - ¾ dy/dx
√3/4 + ¾ dy/dx = -½ - ¾ dy/dx
3/2 dy/dx = -½ - √3/4
dy/dx = -⅓ - √3/(2 x 3)
dy/dx = -⅓ - 1/2√3
(shown)
1/(1 + (xy)²) (y + x dy/dx) = - 1/(1 + x²) - 1/(1 + y²)dy/dx
When x = 1 , y = √3/3
(recall part i's working)
1/(1 + (√3/3)²) (√3/3 + dy/dx) = -1/(1 + 1²) - 1(1 + (√3/3)²) dy/dx
¾(√3/3 + dy/dx)= -1/2 - ¾ dy/dx
√3/4 + ¾ dy/dx = -½ - ¾ dy/dx
3/2 dy/dx = -½ - √3/4
dy/dx = -⅓ - √3/(2 x 3)
dy/dx = -⅓ - 1/2√3
(shown)
Thank you sm!!! :)
See 2 Answers
done
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Good evening Candice! Here is my working for part i.
Date Posted:
4 years ago
done
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Good evening Candice! Here are my workings for parts ii and iii.
Date Posted:
4 years ago
Thank you! :")
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