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How do you differentiate # (2x+1)(x-tanx)#?

Answer 1

Scarlett Allison

#d/(dx)(2x+1)(x-tanx)#

= #4x+1-2xsec^2x-sec^2x-2tanx#

If #f(x)=g(x)xxh(x)#, #(df)/(dx)=g(x)xx(dh)/(dx)+(dg)/(dx)xxh(x)#

Hence #d/(dx)(2x+1)(x-tanx)#

= #(2x+1)(1-sec^2x)+2(x-tanx)#

= #2x+1-2xsec^2x-sec^2x+2x-2tanx#

= #4x+1-2xsec^2x-sec^2x-2tanx#

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Answer 2

Cameron Alexander

To differentiate (2x+1)(x-tanx), you can use the product rule. The product rule states that if you have two functions, u(x) and v(x), then the derivative of their product is given by the formula: (u'v + uv').

Let u(x) = 2x + 1 and v(x) = x - tan(x).

Then, u'(x) = 2 and v'(x) = 1 - sec^2(x) (the derivative of tan(x) is sec^2(x)).

Now, applying the product rule, we have:

(uv)' = (2x + 1)(1 - tan^2(x)) + (x - tan(x))(2)

= (2x + 1)(1 - tan^2(x)) + 2x - 2tan(x)

= 2x + 1 - 2x(tan^2(x)) + 2x - 2tan(x)

= 2 - 2x(tan^2(x)) - 2tan(x)

= 2 - 2x(tan^2(x) + tan(x))

= 2 - 2x(tan(x)(tan(x) + 1))

So, the derivative of (2x+1)(x-tanx) is 2 - 2x(tan(x)(tan(x) + 1)).

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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.