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How do you find the partial derivative of #f(x,y) = sin 3x + cos 5y#?

Answer 1

William Abdalla

Hi there :)

This is the tasteless response.

# f_x(x, y) = 3cos(3x) # and # f_y(x, y) = -5sin(5y) #

The juicy bits are here.

To find # f_x(x, y) #, fix all other variables as constants, in this case just #y#.

So for # f_x(x, y) # we know that #cos(5y)# is only a function of #y# so we differentiate it like a constant since #y# is fixed. So we just differentiate #sin(3x)# in terms of #x# yielding #3cos(3x)#.

This is

# f_x(x, y) = 3cos(3x) #

Similarly, for # f_y(x, y)#, we hold #x# fixed and differentiate with respect to #y# to get

This is

# f_y(x, y) = -5sin(5y) #

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Answer 2

Paisley Johnson

To find the partial derivatives of ( f(x, y) = \sin(3x) + \cos(5y) ):

Partial derivative with respect to ( x ): ( \frac{\partial f}{\partial x} = 3\cos(3x) )
Partial derivative with respect to ( y ): ( \frac{\partial f}{\partial y} = -5\sin(5y) )

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.