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Differentiation of function f(x, y, z) = Sin(x)Sin(y)Sin(z) - Cos(x) Cos(y) Cos(z) w.r.t ‘y’ is?
- a)f’(x, y, z) = Cos(x)Cos(y)Sin(z) + Sin(x)Sin(y)Cos(z)
- b)f’(x, y, z) = Cos(x)Cos(y)Cos(z) + Sin(x)Sin(y)Sin(z)
- c)f’(x, y, z) = Sin(x)Sin(y)Sin(z) + Cos(x)Cos(y)Cos(z)
- d)f’(x, y, z) = Sin(x)Cos(y)Sin(z) + Cos(x)Sin(y)Cos(z)
Correct answer is option 'D'. Can you explain this answer?
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Differentiation of function f(x, y, z) = Sin(x)Sin(y)Sin(z) - Cos(x) C...
f(x, y, z) = Sin(x)Sin(y)Sin(z) - Cos(x) Cos(y) Cos(z)
Since the function has 3 independent variables hence during differentiation we have to consider x and z as constant and differentiate it w.r.t. Y,
f’(x, y, z) = Sin(x)Cos(y)Sin(z) + Cos(x)Sin(y)Cos(z).
Since the function has 3 independent variables hence during differentiation we have to consider x and z as constant and differentiate it w.r.t. Y,
f’(x, y, z) = Sin(x)Cos(y)Sin(z) + Cos(x)Sin(y)Cos(z).
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Differentiation of function f(x, y, z) = Sin(x)Sin(y)Sin(z) - Cos(x) C...
To differentiate the function f(x, y, z) = Sin(x)Sin(y)Sin(z) - Cos(x) Cos(y) Cos(z) with respect to x, we will use the chain rule and the product rule.
Differentiating the first term, Sin(x)Sin(y)Sin(z), with respect to x:
d/dx (Sin(x)Sin(y)Sin(z)) = d/dx (Sin(x)) * Sin(y) * Sin(z)
= Cos(x) * Sin(y) * Sin(z)
Differentiating the second term, -Cos(x) Cos(y) Cos(z), with respect to x:
d/dx (-Cos(x) Cos(y) Cos(z)) = -d/dx (Cos(x)) * Cos(y) * Cos(z)
= -(-Sin(x)) * Cos(y) * Cos(z)
= Sin(x) * Cos(y) * Cos(z)
Now, combining the derivatives of both terms:
d/dx (f(x, y, z)) = d/dx (Sin(x)Sin(y)Sin(z) - Cos(x) Cos(y) Cos(z))
= Cos(x) * Sin(y) * Sin(z) + Sin(x) * Cos(y) * Cos(z)
Therefore, the differentiation of f(x, y, z) with respect to x is Cos(x) * Sin(y) * Sin(z) + Sin(x) * Cos(y) * Cos(z).
Differentiating the first term, Sin(x)Sin(y)Sin(z), with respect to x:
d/dx (Sin(x)Sin(y)Sin(z)) = d/dx (Sin(x)) * Sin(y) * Sin(z)
= Cos(x) * Sin(y) * Sin(z)
Differentiating the second term, -Cos(x) Cos(y) Cos(z), with respect to x:
d/dx (-Cos(x) Cos(y) Cos(z)) = -d/dx (Cos(x)) * Cos(y) * Cos(z)
= -(-Sin(x)) * Cos(y) * Cos(z)
= Sin(x) * Cos(y) * Cos(z)
Now, combining the derivatives of both terms:
d/dx (f(x, y, z)) = d/dx (Sin(x)Sin(y)Sin(z) - Cos(x) Cos(y) Cos(z))
= Cos(x) * Sin(y) * Sin(z) + Sin(x) * Cos(y) * Cos(z)
Therefore, the differentiation of f(x, y, z) with respect to x is Cos(x) * Sin(y) * Sin(z) + Sin(x) * Cos(y) * Cos(z).
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Differentiation of function f(x, y, z) = Sin(x)Sin(y)Sin(z) - Cos(x) Cos(y) Cos(z) w.r.t ‘y’ is?a)f’(x, y, z) = Cos(x)Cos(y)Sin(z) + Sin(x)Sin(y)Cos(z)b)f’(x, y, z) = Cos(x)Cos(y)Cos(z) + Sin(x)Sin(y)Sin(z)c)f’(x, y, z) = Sin(x)Sin(y)Sin(z) + Cos(x)Cos(y)Cos(z)d)f’(x, y, z) = Sin(x)Cos(y)Sin(z) + Cos(x)Sin(y)Cos(z)Correct answer is option 'D'. Can you explain this answer?
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Differentiation of function f(x, y, z) = Sin(x)Sin(y)Sin(z) - Cos(x) Cos(y) Cos(z) w.r.t ‘y’ is?a)f’(x, y, z) = Cos(x)Cos(y)Sin(z) + Sin(x)Sin(y)Cos(z)b)f’(x, y, z) = Cos(x)Cos(y)Cos(z) + Sin(x)Sin(y)Sin(z)c)f’(x, y, z) = Sin(x)Sin(y)Sin(z) + Cos(x)Cos(y)Cos(z)d)f’(x, y, z) = Sin(x)Cos(y)Sin(z) + Cos(x)Sin(y)Cos(z)Correct answer is option 'D'. Can you explain this answer? for Mathematics 2024 is part of Mathematics preparation. The Question and answers have been prepared according to the Mathematics exam syllabus. Information about Differentiation of function f(x, y, z) = Sin(x)Sin(y)Sin(z) - Cos(x) Cos(y) Cos(z) w.r.t ‘y’ is?a)f’(x, y, z) = Cos(x)Cos(y)Sin(z) + Sin(x)Sin(y)Cos(z)b)f’(x, y, z) = Cos(x)Cos(y)Cos(z) + Sin(x)Sin(y)Sin(z)c)f’(x, y, z) = Sin(x)Sin(y)Sin(z) + Cos(x)Cos(y)Cos(z)d)f’(x, y, z) = Sin(x)Cos(y)Sin(z) + Cos(x)Sin(y)Cos(z)Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Mathematics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Differentiation of function f(x, y, z) = Sin(x)Sin(y)Sin(z) - Cos(x) Cos(y) Cos(z) w.r.t ‘y’ is?a)f’(x, y, z) = Cos(x)Cos(y)Sin(z) + Sin(x)Sin(y)Cos(z)b)f’(x, y, z) = Cos(x)Cos(y)Cos(z) + Sin(x)Sin(y)Sin(z)c)f’(x, y, z) = Sin(x)Sin(y)Sin(z) + Cos(x)Cos(y)Cos(z)d)f’(x, y, z) = Sin(x)Cos(y)Sin(z) + Cos(x)Sin(y)Cos(z)Correct answer is option 'D'. Can you explain this answer?.
Differentiation of function f(x, y, z) = Sin(x)Sin(y)Sin(z) - Cos(x) Cos(y) Cos(z) w.r.t ‘y’ is?a)f’(x, y, z) = Cos(x)Cos(y)Sin(z) + Sin(x)Sin(y)Cos(z)b)f’(x, y, z) = Cos(x)Cos(y)Cos(z) + Sin(x)Sin(y)Sin(z)c)f’(x, y, z) = Sin(x)Sin(y)Sin(z) + Cos(x)Cos(y)Cos(z)d)f’(x, y, z) = Sin(x)Cos(y)Sin(z) + Cos(x)Sin(y)Cos(z)Correct answer is option 'D'. Can you explain this answer? for Mathematics 2024 is part of Mathematics preparation. The Question and answers have been prepared according to the Mathematics exam syllabus. Information about Differentiation of function f(x, y, z) = Sin(x)Sin(y)Sin(z) - Cos(x) Cos(y) Cos(z) w.r.t ‘y’ is?a)f’(x, y, z) = Cos(x)Cos(y)Sin(z) + Sin(x)Sin(y)Cos(z)b)f’(x, y, z) = Cos(x)Cos(y)Cos(z) + Sin(x)Sin(y)Sin(z)c)f’(x, y, z) = Sin(x)Sin(y)Sin(z) + Cos(x)Cos(y)Cos(z)d)f’(x, y, z) = Sin(x)Cos(y)Sin(z) + Cos(x)Sin(y)Cos(z)Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Mathematics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Differentiation of function f(x, y, z) = Sin(x)Sin(y)Sin(z) - Cos(x) Cos(y) Cos(z) w.r.t ‘y’ is?a)f’(x, y, z) = Cos(x)Cos(y)Sin(z) + Sin(x)Sin(y)Cos(z)b)f’(x, y, z) = Cos(x)Cos(y)Cos(z) + Sin(x)Sin(y)Sin(z)c)f’(x, y, z) = Sin(x)Sin(y)Sin(z) + Cos(x)Cos(y)Cos(z)d)f’(x, y, z) = Sin(x)Cos(y)Sin(z) + Cos(x)Sin(y)Cos(z)Correct answer is option 'D'. Can you explain this answer?.
Solutions for Differentiation of function f(x, y, z) = Sin(x)Sin(y)Sin(z) - Cos(x) Cos(y) Cos(z) w.r.t ‘y’ is?a)f’(x, y, z) = Cos(x)Cos(y)Sin(z) + Sin(x)Sin(y)Cos(z)b)f’(x, y, z) = Cos(x)Cos(y)Cos(z) + Sin(x)Sin(y)Sin(z)c)f’(x, y, z) = Sin(x)Sin(y)Sin(z) + Cos(x)Cos(y)Cos(z)d)f’(x, y, z) = Sin(x)Cos(y)Sin(z) + Cos(x)Sin(y)Cos(z)Correct answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for Mathematics.
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Differentiation of function f(x, y, z) = Sin(x)Sin(y)Sin(z) - Cos(x) Cos(y) Cos(z) w.r.t ‘y’ is?a)f’(x, y, z) = Cos(x)Cos(y)Sin(z) + Sin(x)Sin(y)Cos(z)b)f’(x, y, z) = Cos(x)Cos(y)Cos(z) + Sin(x)Sin(y)Sin(z)c)f’(x, y, z) = Sin(x)Sin(y)Sin(z) + Cos(x)Cos(y)Cos(z)d)f’(x, y, z) = Sin(x)Cos(y)Sin(z) + Cos(x)Sin(y)Cos(z)Correct answer is option 'D'. Can you explain this answer?, a detailed solution for Differentiation of function f(x, y, z) = Sin(x)Sin(y)Sin(z) - Cos(x) Cos(y) Cos(z) w.r.t ‘y’ is?a)f’(x, y, z) = Cos(x)Cos(y)Sin(z) + Sin(x)Sin(y)Cos(z)b)f’(x, y, z) = Cos(x)Cos(y)Cos(z) + Sin(x)Sin(y)Sin(z)c)f’(x, y, z) = Sin(x)Sin(y)Sin(z) + Cos(x)Cos(y)Cos(z)d)f’(x, y, z) = Sin(x)Cos(y)Sin(z) + Cos(x)Sin(y)Cos(z)Correct answer is option 'D'. Can you explain this answer? has been provided alongside types of Differentiation of function f(x, y, z) = Sin(x)Sin(y)Sin(z) - Cos(x) Cos(y) Cos(z) w.r.t ‘y’ is?a)f’(x, y, z) = Cos(x)Cos(y)Sin(z) + Sin(x)Sin(y)Cos(z)b)f’(x, y, z) = Cos(x)Cos(y)Cos(z) + Sin(x)Sin(y)Sin(z)c)f’(x, y, z) = Sin(x)Sin(y)Sin(z) + Cos(x)Cos(y)Cos(z)d)f’(x, y, z) = Sin(x)Cos(y)Sin(z) + Cos(x)Sin(y)Cos(z)Correct answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
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