Find the partial derivatives of the following: $f(x,y,z)=x^{\sin(y^{x})}+\int_{0}^{x} t^tdt$.

Question

$f(x,y,z)=x^{\sin(y^{x})}+\int_{0}^{x} t^tdt$.

Im not sure how to treat this integral in relation to the different variables.. and the first part also is unclear. $x^{\sin(y^{x})}$ :D

Answer 1

$ x^{\sin(y^x)} = \exp(\ln(x)\cdot \sin(y^x)))$. Also, the fundamental theorem of calculus gives you $\frac{\partial}{\partial x}(\int_{0}^{x} t^t dt) = x^x$. For finding the derivative of $\sin(y^x)$, we can again write $\sin(y^x) = \sin(\exp(x\cdot\ln(y)))$ and apply the chain rule. By doing this we obtain

$\frac{\partial}{\partial x} \sin(y^x) = y^x \cdot \cos(y^x) \cdot \ln(y)$.

Using this we get $\frac{\partial f}{\partial x} = x^{\sin(y^x)} \cdot (\frac{\sin(y^x)}{x} + y^x \cdot \cos(y^x) \cdot \ln(y)) + x^x$

Using the same technique we get

$\frac{\partial}{\partial y}(\sin(y^x)) = xy^{x-1}\cos(y^x)$. Because the integral does not depend on $y$, its derivative with respect to $y$ is zero, and we obtain

$\frac{\partial f}{\partial y} = x^{\sin(y^x)+1}y^{x-1}\cos(y^x)\ln(x)$

Clearly, $\frac{\partial f}{\partial z} = 0$.

Answer 2

Suppose $\;F(x)\;$ is s.t. $\;F'(x)=x^x\;,\;\;x>0\;$ , so

$$f(x,y,z)=e^{\sin y^x}+F(x)-F(0)\implies\;\;\text{for example:}$$

$$f'_x=e^{\sin y^x}y^x\cos y^x\log y+F'(x)=e^{\sin y^x}y^x\cos y^x\log y+x^x$$

and etc.

Answer 3

This integral depends only of x so: $$\dfrac{\partial \int_0^x t^t }{\partial x}= F`(x)-F`(0)=x^x$$ where $F$ is the primitive of $t^t$ $$\dfrac{\partial \int_0^x t^t }{\partial y} = 0$$ and $$\dfrac{\partial \int_0^x t^t }{\partial z} = 0$$

(I'm sorry, was writing when Timbuc posted)