# partial derivative of definite intergral

Question: Find the partial derivatives, $$f_x(x, y)$$ and $$f_y(x,y)$$, of the function $$f(x,y)=\int_y^xcos(3t^2+9t-1)dt$$

My attempt is as follows.

1. Substitution:

$$u=3t^2+9t-1$$

$$\frac{du}{dt}=6t+9$$

$$dt=\frac{1}{6t+9}du$$

2. Plug in $$u$$:

$$\int_y^xcos(3t^2+9t-1)dt$$

$$=\int_y^x\frac{cos(u)}{6t+9}du$$

and I don't know how to continue.

I've been stuck on this question for a few days now. Tried searching across all platforms but none has similar questions like this.

WolframAlpha shows an answer that involves the Fresnel C and S integrals but my class hasn't mentioned this anywhere.

• Just use the Leibniz integral rule. You should get $$f_{x} = \cos(3x^{2}+9x-1)$$ and $$f_{y} = -\cos(3y^{2}+9y-1)$$ – mattos Apr 13 '19 at 4:54

## 1 Answer

No need to calculate the integral: $$f_(x,y)=\cos(3x^{2}+9x-1)$$ and $$f_y((x,y)=-\cos(3y^{2}+9y-1)$$. You only have to know that the derivative of an indefinite integral gives back the original function.

• Oh thank you so much! So is it a general solution that $f_y (x,y)$, the lower limit, is just the original function with a negative sign? – Vero Apr 13 '19 at 8:01