How do you find the derivative of the function: #arctan (cos x)#?

Answer 1

#(-sinx)/(1+cos^2x)#

differentiate using the #color(blue)"chain rule"#
#color(red)(|bar(ul(color(white)(a/a)color(black)(dy/dx=(dy)/(du)xx(du)/(dx))color(white)(a/a)|)))........ (A)#
#color(orange)"Reminder " color(red)(|bar(ul(color(white)(a/a)color(black)(d/dx(arctanx)=1/(1+x^2))color(white)(a/a)|)))#
let #u=cosxrArr(du)/(dx)=-sinx#
and #y=arctanurArr(dy)/(du)=1/(1+u^2)#

substitute these values into (A) changing u back to x.

#rArrdy/dx=1/(1+u^2)xx(-sinx)=(-sinx)/(1+cos^2x)#
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Answer 2

To find the derivative of the function arctan(cos(x)), you apply the chain rule.

Let y = arctan(cos(x)).

Now, take the derivative of both sides with respect to x:

dy/dx = d(arctan(cos(x)))/dx.

By the chain rule, d(arctan(u))/du = 1/(1 + u^2), where u = cos(x).

So, d(arctan(cos(x)))/dx = -sin(x)/(1 + cos^2(x)).

Therefore, the derivative of arctan(cos(x)) with respect to x is -sin(x)/(1 + cos^2(x)).

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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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