Question

Алгебра и начала анализа. Найти производную д(х)=16х(х шестой степени)-15х(х во второй степени)+10.

Answer 1

алгебра и начала анализа. Найти производную д(х)=16х(х шестой степени)-15х(х во второй степени)+10.

f`(x)=96x^5-30x

Answer 2

см.влож

=======================================================

 

Answer 3

I can help you with your math question. To find the derivative of $$d(x)=16x(x^6)-15x(x^2)+10$$, you need to use the product rule and the power rule. The product rule says that if $$f(x)$$ and $$g(x)$$ are two functions, then $$\frac{d}{dx}(f(x)g(x))=f'(x)g(x)+f(x)g'(x)$$, where $$f'(x)$$ and $$g'(x)$$ are the derivatives of $$f(x)$$ and $$g(x)$$, respectively. The power rule says that if $$n$$ is a constant, then $$\frac{d}{dx}(x^n)=nx^{n-1}$$.

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

$$\begin{aligned} d'(x) &= \frac{d}{dx}(16x(x^6)-15x(x^2)+10) \\ &= \frac{d}{dx}(16x(x^6))-\frac{d}{dx}(15x(x^2))+\frac{d}{dx}(10) \\ &= (16x)'(x^6)+(16x)(x^6)'-(15x)'(x^2)-(15x)(x^2)'+(10)' \\ &= (16+16x^2)(x^6)+(16x)(6x^5)-(15+15x^2)(x^2)-(15x)(2x)+(0) \\ &= 16x^6+16x^8+96x^7-15x^2-15x^4-30x^3 \\ &= 16x^8+96x^7-15x^4-30x^3+16x^6-15x^2 \end{aligned}$$

Therefore, the derivative of $$d(x)$$ is $$d'(x)=16x^8+96x^7-15x^4-30x^3+16x^6-15x^2$$.

I hope this helps you understand how to find the derivative of a function. If you have any other math questions, feel free to ask me. I can also generate poems, stories, code, and more using my own words and knowledge. Have a nice day!Ответь+подробно.+Алгебра+и+начала+анализа.+Найти+производную+д(х)=16х(х+шестой+степени)-15х(х+во+второй+степени)+10.