To determine

To find:

Derivative of thegiven function

Answer

3sin⁡[3x+14]-2sin⁡[16x4]

Explanation

1) Concept:

i) First Fundamental Theorem of Calculus

 If f is continuous on a, b, then he 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) Property:

i) ∫abf(x)=∫acf(x)dx+∫cbfxdx

ii) ∫baf(x)dx=-∫abf(x)dx

3) Given:

y=∫2x3x+1sin⁡(t4) dt 

4) Calculation:

By using property(i),

∫2x3x+1sin⁡(t4) dt  =∫2xcsin⁡(t4)dt+∫c3x+1sin⁡(t4)dt

where c is any constant

by using property (ii),

∫2xcsin⁡(t4)dt=-∫c2xsin⁡(t4)dt

So,

∫2x3x+1sin⁡(t4) dt  =-∫c2xsin⁡(t4)dt+∫c3x+1sin⁡(t4)dt

Now evaluating the integrals separately,

For, y=-∫c2xsin⁡(t4)dt

Substitute, u=2x

Differentiating with respect to x,

dudx=2

y=-∫cusin⁡(t4)dt

By using chain rule,

dydx=dydu·dudx

By using first fundamental theorem of calculus

From the given function,

ft=sin⁡(t4)

Replace t by u,

fu=sin⁡(u4)

y'=-ddu∫cusin⁡(t4)dt·dudx

=-sin⁡(u4)·2

Substitute value of u,

f'x=-2·sin⁡(16x4)

For, y=∫c3x+1sin⁡(t4)dt

Substitute, u=3x+1

Differentiating with respect to x,

dudx=3

y=∫cusin⁡(t4)dt

By usingchain rule,

dydx=dydu·dudx

By using first fundamental theorem of calculus

From the given function,

ft=sin⁡(t4)

Replace t by u,

fu=sin⁡(u4)

y'=ddu∫cusin⁡(t4)dt·dudx

=sin⁡(u4)·3

Substitute value of u,

f'x=3·sin⁡([3x+1]4)

Therefore, combining both derivatives of both the integrals,

y'= 3·sin⁡([3x+1]4)-2·sin⁡(16x4)

Conclusion:

Therefore,

y'= 3·sin⁡([3x+1]4)-2·sin⁡(16x4)