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If y equals the integral from 5x on top to cosx on bottom of cos of u squared du what is y'?
Wiki User
∙ 15y ago
y=S^5x _cos(x) cos(u²) du
The derivative of a definite integral of a function f(x) is equal to the difference in the product of the function at each limit of integration times the limit of integration.
y'=cos(u²)*du/dx from u=cos(x) to u=5x
y'=-sin(x)*cos(cos(x)²)-5*cos(25x²)
To understand why this works, consider the following where F(x) is the antiderivative of f(x)
y=F(g(x))-F(h(x))=S f(x)dx from h(x) to g(x)
If you take the derivative of this expression and apply the chain rule
dy/dx = dF(g(x))/dx - dF(h(x))/dx = f(g(x))*dg/dx - f(h(x))*dh/dx
Wiki User
∙ 15y ago
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