To determine
To find:
Derivative of thegiven function
Answer
g'x=cos3x1+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=∫1sinx1-t21+t4 dt
3) Calculation:
Substitute u=sinx
Differentiating with respect to x,
dudx=cosx
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·cosx
Substitute value of u,
g'x=1-sin2x1+sin4x·cosx
Since, 1-sin2x=cos2x
=cos2x1+sin4x·cosx
=cos3x1+sin4x
Hence derivative of the function is cos3x1+sin4x
Conclusion:
Therefore,
g'x=cos3x1+sin4x