Differentiate #sinx# #/# #5x# + #sec^2 x"# ?

Answer 1

#(xcosx-sinx)/(5x^2)+2sec^2xtanx#

#d/dx(sinx/(5x)+sec^2x)# #=d/dx(sinx/(5x))+d/dx(sec^2x)# #=((d(sinx))/dx(5x)-sinx(d(5x))/dx)/(5x)^2+(d(sec^2x))/(dsecx)*(d(secx))/dx# #=(cosx*5x-5sinx)/(25x^2)+2secx*secxtanx#
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Answer 2

To differentiate the function ( \frac{\sin(x)}{5x + \sec^2(x)} ), you can use the quotient rule of differentiation, which states that if you have a function ( \frac{f(x)}{g(x)} ), then its derivative is given by:

[ \frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} ]

So, applying the quotient rule to the given function:

[ \frac{d}{dx}\left(\frac{\sin(x)}{5x + \sec^2(x)}\right) = \frac{(5x + \sec^2(x))(\cos(x)) - \sin(x)(5 + 2\sec(x)\tan(x))}{(5x + \sec^2(x))^2} ]

This gives you the derivative of the given function.

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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.

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.

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.

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.

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