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How do you differentiate #f(x)=tan(1-3x) # using the chain rule?

Answer 1

Ezra Adams

#f'(x)=-3sec^2(1-3x)#

Using the general rule #d/dxtan[u(x)]=sec^2u*(du)/dx#,

we get #d/dxtan(1-3x)=-3sec^2(1-3x)#

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Answer 2

Ezra Adams

# - 3 sec^2(1 - 3x ) #

differentiating using the 'chain rule ' :

# f'(x) = sec^2(1 - 3x ) . d/dx (1 - 3x ) #

# = sec^2 (1 - 3x ) .(- 3 ) = - 3 sec^2 (1 - 3x ) #

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Answer 3

Axel Alvarez

To differentiate ( f(x) = \tan(1 - 3x) ) using the chain rule, follow these steps:

Identify the outer function (( \tan(x) )) and the inner function (( 1 - 3x )).
Differentiate the outer function with respect to the inner function.
Differentiate the inner function with respect to ( x ).
Multiply the results from steps 2 and 3 to obtain the derivative.

The derivative of ( f(x) ) is:

[ f'(x) = \frac{d}{dx}[\tan(1 - 3x)] = \frac{d}{d(1 - 3x)}[\tan(x)] \cdot \frac{d}{dx}(1 - 3x) ]

Now, differentiate each part:

[ \frac{d}{d(1 - 3x)}[\tan(x)] = \sec^2(1 - 3x) ] [ \frac{d}{dx}(1 - 3x) = -3 ]

Multiply these results:

[ f'(x) = -3 \sec^2(1 - 3x) ]

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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.