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I know this is probably a really easy question. However I keep reading the examples again and again and constantly cant get the correct answer.

Find the value of $\frac{dy}{dx}$ when $x=1$ $$ y = a^2x -ax^2$$

Is is possible someone could do a detailed step by step solution. Explain it like I am 5?

kmeis
  • 550
allquiet1984
  • 149

1 Answers1

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Assuming $a$ is a constant then:

$$y'(x) = \frac{d}{dx}(a^2x - ax^2) = \frac{d}{dx}(a^2x) - \frac{d}{dx}(ax^2)= a^2\frac{d}{dx}(x) - a\frac{d}{dx}(x^2) = a^2 - 2ax$$

Then evaluate at $x=1$:

$$y'(1) = a^2-2a$$

kmeis
  • 550
  • Thank you so much for that! It really helped. I thought I was meant to sub it in first. I feel so stupid. – allquiet1984 Jan 29 '17 at 23:51
  • Its no trouble. It is a common error to substitute first then derive. However, consider that for a function $y(x)$ substituting in first then $y(1)$ will be a number. The derivative of a constant function is zero. – kmeis Jan 29 '17 at 23:59