How do you find the partial derivative of the function $f (x, y) = \int cos (- 7 t^{2} - 6 t - 1) d t$ $f(x,y)=intcos(-7t^2-6t-1)dt$ ?

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How do you find the partial derivative of the function $f (x, y) = \int cos (- 7 t^{2} - 6 t - 1) d t$ $f(x,y)=intcos(-7t^2-6t-1)dt$ ?

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Answer 1 · 2015-03-15T17:24:56

You need to be more specific about how $f$ $f$ depends on $x$ $x$ and $y$ $y$ . Are they in the limits of the integral? Are they part of the integrand function? As you wrote it, there's no $x$ $x$ or $y$ $y$ dependence on the right-hand side and the derivatives with respect to $x$ $x$ and $y$ $y$ would be zero.

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How do you find the partial derivative of the function $f (x, y) = \int cos (- 7 t^{2} - 6 t - 1) d t$ $f(x,y)=intcos(-7t^2-6t-1)dt$ ?

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How do you find the partial derivative of the function f(x,y)=∫cos(−7t2−6t−1)dtf(x,y)=intcos(-7t^2-6t-1)dt?

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How do you find the partial derivative of the function $f (x, y) = \int cos (- 7 t^{2} - 6 t - 1) d t$ $f(x,y)=intcos(-7t^2-6t-1)dt$ ?