# Calculating Derivative – Deriving a function in another function – Exercise 6279

Exercise

given the following function:

$$y=\frac{x}{\sqrt{9-x^2}}-f(\frac{x}{3})$$

The following holds:

$$f'(x)=\frac{1}{\sqrt{1-x^2}}$$

Find the derivative of y.

$$y'=\frac{x^2}{\sqrt{{(9-x^2)}^3}}$$

Solution

$$y=\frac{x}{\sqrt{9-x^2}}-f(\frac{x}{3})$$

Using Derivative formulas and the quotient rule and chain rule in Derivative Rules, we get the derivative:

$$y'=\frac{\sqrt{9-x^2}-x\cdot\frac{1}{2\sqrt{9-x^2}}\cdot (-2x)}{9-x^2}-\frac{1}{\sqrt{1-{(\frac{x}{3})}^2}}\cdot \frac{1}{3}=$$

הערה: השתמשנו בנגזרת של f הנתונה בשאלה וכפלנו בנגזרת הפנימית לפי כלל ההרכבה.

Note: We used the derivative of f given in the question and multiplied the internal derivative by the chain rule in Derivative Rules.

We simplify the derivative:

$$y'=\frac{\sqrt{9-x^2}+\frac{2x^2}{2\sqrt{9-x^2}}}{9-x^2}-\frac{1}{\sqrt{1-\frac{x^2}{9}}}\cdot \frac{1}{3}=$$

$$=\frac{\frac{2(9-x^2)+2x^2}{2\sqrt{9-x^2}}}{9-x^2}-\frac{1}{3\sqrt{\frac{9-x^2}{9}}}=$$

$$=\frac{9-x^2+x^2}{(9-x^2)\sqrt{9-x^2}}-\frac{1}{\sqrt{9-x^2}}=$$

$$=\frac{9-(9-x^2)}{\sqrt{{(9-x^2)}^3}}=$$

$$=\frac{x^2}{\sqrt{{(9-x^2)}^3}}$$

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