ⅆⅆx(∫x2sin(t4)ⅆt)=
sin(x4) – x to the fourth
To solve this problem, we need to apply the chain rule and the fundamental theorem of calculus.
Firstly, let’s simplify the integrand with respect to t using substitution method.
Let u = t^4
Then, du/dt = 4t^3
and dt = du/4t^3
Substituting in the integral:
∫x^2 sin(t^4) dt = (∫sin(u)du)/4t^3
= (-cos(u))/4t^3
= (-cos(t^4))/4t^3
Now, let’s apply the chain rule to differentiate the expression with respect to x:
ⅆⅆx(∫x^2 sin(t^4) dt)
= ∫ⅆⅆx (-cos(t^4))/4t^3 ⅆt
= ∫(-cos(t^4))/4t^3 (ⅆt/ⅆx) ⅆt
= ∫(-cos(t^4))/4t^3 (ⅆ(t^4)/ⅆx) ⅆt
= ∫-4x^3 cos(t^4)/4t^3 ⅆt
= -x^3 sin(t^4) + C
Hence, the final result is -x^3 sin(t^4) + C.
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