To determine

To find:

Derivative of thegiven function

Answer

g'x=cos3⁡x1+sin4x

Explanation

1) Concept:

i) First Fundamental Theorem of Calculus

 If f is continuous on a, b, then function g defined by

gx=∫axf(t)dt

is continuous on a, b and differentiable on (a, b), and g'x=f(x)

ii) Chain Rule:

If y=f(u) and u=g(x) are both differentiable functions, then

dydu=dydu·dudx

2) Given:

gx=∫1sin⁡x1-t21+t4 dt 

3) Calculation:

Substitute u=sin⁡x

Differentiating with respect to x,

dudx=cos⁡x

gx=∫1u1-t21+t4 dt 

By use of chain rule,

dgdx=dgdu·dudx

By using first fundamental theorem of calculus

From the given function,

ft=1-t21+t4

Replace t by u,

fu=1-u21+u4

g'(x)=ddu∫0u1-t21+t4 dt ·dudx

g'x=fx=1-u21+u4·cos⁡x

Substitute value of u,

g'x=1-sin2x1+sin4x·cos⁡x

Since, 1-sin2x=cos2⁡x

=cos2⁡x1+sin4x·cos⁡x

=cos3⁡x1+sin4x

Hence derivative of the function is cos3⁡x1+sin4x

Conclusion:

Therefore,

g'x=cos3⁡x1+sin4x