How do you differentiate #f(x)=x^2lnsec(x^2)# using the chain rule?

Answer 1

#dy/dx =2x(x^2*tan(x^2) + ln(sec(x^2)))#

So we have

#y = x^2ln(sec(x^2))#

Derivating and using the product rule we have

#dy/dx = x^2d/dx(ln(sec(x^2)) + ln(sec(x^2))d/dxx^2#
#dy/dx = x^2d/dx(ln(sec(x^2)) + 2xln(sec(x^2))#
Let's say #sec(x^2) = u# then, by the chain rule we have
#dy/dx = x^2d/dx(ln(u)) + 2xln(sec(x^2))# #dy/dx = x^2d/(du)(ln(u))(du)/dx + 2xln(sec(x^2))# #dy/dx = x^2/u*d/dx(sec(x^2)) + 2xln(sec(x^2))#
Let's say #x^2 = v# then we have
#dy/dx = x^2/u*d/(dv)sec(v)*(dv)/dx + 2xln(sec(x^2))#
#dy/dx = x^2/u*tan(v)*sec(v)*2x + 2xln(sec(x^2))#
Putting it all in terms of #x# we have
#dy/dx =2x(x^2/sec(x^2)*tan(x^2)*sec(x^2) + ln(sec(x^2)))#
#dy/dx =2x(x^2*tan(x^2) + ln(sec(x^2)))#
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Answer 2

To differentiate the function ( f(x) = x^2 \ln(\sec(x^2)) ) using the chain rule, follow these steps:

  1. Identify the outer function ( u ) and the inner function ( v ):

    • Outer function ( u ): ( \ln(u) )
    • Inner function ( v ): ( \sec(v) )
  2. Differentiate the outer function with respect to its argument:

    • ( \frac{d}{dx}(\ln(u)) = \frac{1}{u} \frac{du}{dx} )
  3. Differentiate the inner function with respect to ( x ):

    • ( \frac{d}{dx}(\sec(v)) = \sec(v) \tan(v) \frac{dv}{dx} )
  4. Substitute the expressions for ( u ) and ( v ) into the derivative of the outer and inner functions.

  5. Apply the chain rule by multiplying the derivative of the outer function with the derivative of the inner function.

  6. Simplify the expression obtained from step 5.

  7. Finally, substitute the expression for ( \frac{dv}{dx} ) into the simplified result to get the final derivative.

The derivative of the function ( f(x) = x^2 \ln(\sec(x^2)) ) using the chain rule is:

[ f'(x) = x^2 \left( \frac{2}{x} \ln(\sec(x^2)) + \frac{1}{\sec(x^2)} \cdot \sec(x^2) \tan(x^2) \cdot 2x \right) ]

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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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