How do you differentiate #y= x+((x+sin^2x)^3) ^4#?

Answer 1

#y' = 1+12(x + sin^2(x))^11 (1-2sin(x)cos(x))#

This problem is solved using the chain rule: #d/dx f(g(x)) = f'(g(x))*g'(x)#
#y = x + ((x + sin^2(x))^3)^4 = x + (x+sin^2(x))^12#
Taking the derivative: #(dy)/dx = d/dx x + d/dx (x+sin^2(x))^12# #= 1 + 12(x + sin^2(x))^11*(d/dx(x + sin^2(x)))# #= 1 + 12(x + sin^2(x))^11*(d/dx x + d/dx sin^2(x))# #= 1 + 12(x + sin^2(x))^11*(1 + 2sin(x)(d/dx sin(x)))# #= 1 + 12(x + sin^2(x))^11(1 - 2sin(x)cos(x))#
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Answer 2

To differentiate the given function y = x + ((x + sin^2x)^3)^4, you can apply the chain rule and the power rule.

Let's denote the inner function as u(x) = (x + sin^2x)^3.

Now, differentiate u(x) with respect to x:

u'(x) = 3(x + sin^2x)^2 * (1 + 2sin(x)cos(x))

Now, let's denote the outer function as v(u) = u^4.

Differentiate v(u) with respect to u:

v'(u) = 4u^3

Now, apply the chain rule:

y' = v'(u) * u'(x)

Substitute u(x) and its derivative into the equation:

y' = 4(x + sin^2x)^3 * (1 + 2sin(x)cos(x)) * (x + sin^2x)^2

Simplify the expression:

y' = 4(x + sin^2x)^5(1 + 2sin(x)cos(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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