What is the derivative of #sinx(tanx)#?

Answer 1

#sinx(sec^2x+1)#

Differentiate using the #color(blue)"product rule"#
#color(red)(|bar(ul(color(white)(a/a)color(black)(f(x)=g(x)h(x)" then" f'(x)=g(x)h'(x)+h(x)g'(x))color(white)(a/a)|)))#
#g(x)=sinxrArrg'(x)=cosx#
#h(x)=tanxrArrh'(x)=sec^2x# #"--------------------------------------------------"# Substitute these values into f'(x)
#rArrf'(x)=sinxsec^2x+tanxcosx#
Now #tanxcosx=sinx/cancel(cosx)xxcancel(cosx)=sinx#
#rArrf'(x)=sinxsec^2x+sinx=sinx(sec^2x+1)#
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Answer 2

To find the derivative of sin(x)tan(x), we can use the product rule of differentiation, which states that the derivative of a product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function.

Applying the product rule, we have:

d/dx [sin(x)tan(x)] = (cos(x)tan(x) + sin(x)sec^2(x))

So, the derivative of sin(x)tan(x) with respect to x is cos(x)tan(x) + sin(x)sec^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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