- -2/x. The derivative of f(g(x)) is just that of f(x) multiplied by that of g(x). The derivative of x2 is simply 2x. That of 1/x, or x-1, is less well known. We multiply by the exponent and then decrease the exponent by 1. We get -x-2, or -1/x2. 2x * -1/x2 = -2/x.
- Pi. The area of an ellipse is the minor radius times the major radius times pi. This means f(x) is equivalent to πx when x is more than 1. Otherwise it is undefined. And the derivative of any number multiplied by x is just that number! Our answer is π.
- -1/(x+√x)2. We simply multiply by the exponent and then decrease the exponent by 1. It’s the same as #1, don’t let the all the fancy stuff lead you astray. -(x+√x)-2 or -1/(x+√x)2 is our answer.
- ((cx2+cx)/(x2-2))+c. Before evaluating it’s antiderivative, we can multiply both the top and bottom of the fraction by x-√2, which preserves it’s value. On the bottom, (x+√2)*(x-√2) equals x2-2. And on the top, we have c(x-√2) = cx-x. Thus we can simplify the problem to the antiderivative of cx-x / x2-2. This is simply the antiderivative of cx-c divided by x2-2 plus a constant. We get ((cx2+cx) / (x2-2))+c.
- -0.5±√(x2+c)/4(x2+c). First let’s use the quadratic formula. n equals -x±√(x2+c) divided by 2. This means our equation is equivalent to half the derivative of -x±√(x2+c). This in turn equals half the derivative of -x, -0.5, plus or minus the derivative of √(x2+c). For the second part, remember the rule in #1. We get √(x2+c) / 2(x2+c). Our answer is thus -0.5±√(x2+c)/4(x2+c).
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Challenges cover a variety of topics