# Find the derivative of: `int e^(sint) dt` , `[cosx, x^2]`

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### 1 Answer

Apply the second fundamental theorem of calculus which is:

`F'(x) = d/(dx)int_a^x f(t)dt = f(x)`

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Given the interval `[cosx, x^2]` ` `, let the boundaries of the interval be the lower and upper limit.

`F'(x) = d/(dx)int_(cosx)^(x^2) e^(sint)dt`

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Then, let's take a point on the given interval. Let this point be `a` ` `. Then, re-write the integral with the following limits:

`= d/(dx)[int_(cosx)^ae^(sint)dt + int_a^(x^2)e^(sint)dt]=d/(dx)[-int_a^(cosx)e^(sint)dt + int_a^(x^2) e^(sint)dt]`

`=d/(dx)[int_a^(x^2)e^(sint)dt-int_a^(cosx)e^(sint)dt]`

Let's take the derivative of each integral separately.

> Let `h'(x)=d/(dx)int_a^(x^2)e^(sint)dt` ` ` . To take the derivative of h, apply chain rule. So we have,

` `` u=x^2 ` and `h=int_a^ue^(sint)dt `

`(du)/(dx)=2x` `(dh)/(du)= d/(du)int_a^ue^(sint)dt`

To take the derivative of h with respect to u, apply the Second Fundamental Theorem of Calculus.

`(dh)/(du)=d/(du)int_a^ue^(sint)dt=e^(sinu)`

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Then, multiply the two derivatives to get h'(x).

`h'(x)=(du)/(dx)*(dh)/(du)=2x*e^(sinu)`

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Susbtitute `u=x^2` .

`h'(x)=2x*e^(sinx^2)`

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> Let `g'(x)=d/(dx)int_a^(cosx)e^(sint)dt` ` ` . Use the chain rule of derivatives to determine g'(x).

` ` `v=cosx ` and `g=int_a^ve^(sint)dt `

`(dv)/(dx)=-sinx ` `(dg)/(dv)=d/(dv)int_a^ve^(sint)tdt`

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To solve for derivative of g with respect to v, use the second fundametal theorem of calculus.

`(dg)/(dv)=d/(dv)int_a^ve^(sint)tdt = e^(sinv) = e^(sin(cosx))`

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Multiply the two derivatives to get g'(x).

`g'(x)=(dv)/(dx)*(dg)/(dv)=-sinx*e^(sin(cosx))`

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Next, subtract h'(x) and g'(x) to determine F'(x).

`F'(x) = d/(dx)int_a^(x^2)e^(sint)dt - d/(dx)int_a^(cosx)e^(sint)dt = h'(x)-g'(x)`

`F'(x)=2x*e^(sinx^2) - (-sinx*e^[sin(cosx)]) = 2x*e^(sinx^2)+sinx*e^[sin(cosx)]`

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**Hence, derivative of `inte^(sint)dt ` on the interval `[cosx,x^2]` is `2x*e^(sinx^2)+sinx*e^[sin(cosx)]` .**