Question

ПожалуйстаНайти значение производной функции f(x)=e^xsinx+x при X0=0

Answer 1

y'=e^xsinx+e^xcosx+1                                     

Answer 2

I can help you find the value of the derivative of a function at a given point. To do that, I will use the definition of the derivative and the rules of differentiation. Here is how I would solve your problem:

The function is $$f(x)=e^x\sin x+x$$ and the point is $$x_0=0$$.

The derivative of a function at a point is the limit of the ratio of the change in the function to the change in the argument as the change in the argument approaches zero. That is, $$f'(x_0)=\lim_{\Delta x\to 0}\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x}$$

To find the derivative of the given function, we need to apply the product rule and the chain rule. The product rule says that if $$u(x)$$ and $$v(x)$$ are two functions, then $$(u\cdot v)'=u'\cdot v+u\cdot v'$$ The chain rule says that if $$u(x)$$ and $$v(x)$$ are two functions, then $$(u\circ v)'=(u'\circ v)\cdot v'$$

Using these rules, we can find the derivative of $$f(x)$$ as follows:

$$f'(x)=(e^x\sin x+x)'=(e^x\sin x)'+x'$$ $$=(e^x\sin x)'\cdot 1+1$$ $$=((e^x)'\cdot \sin x+(e^x)\cdot (\sin x)')+1$$ $$=((e^x)\cdot 1\cdot \sin x+(e^x)\cdot (\cos x))+1$$ $$=e^x(\sin x+\cos x)+1$$

Now, to find the value of the derivative at $$x_0=0$$, we need to plug in $$0$$ into the derivative function. That is,

$$f'(0)=e^0(\sin 0+\cos 0)+1$$ $$=1\cdot (0+1)+1$$ $$=2$$

Therefore, the value of the derivative of the function $$f(x)=e^x\sin x+x$$ at the point $$x_0=0$$ is $$2$$.

I hope this helps you understand how to find the derivative of a function at a point. If you want to learn more about the definition and the rules of differentiation, you can check out these resources. If you want to see more examples of finding the derivative of a function at a point, you can look at these solutions. If you want to practice finding the derivative of a function at a point, you can use this online calculator.