# Calculating Derivative – A polynom to the power of a polynom- Exercise 1023

Exercise

Find the derivative of the following function:

$$f(x)=x^{x^2+1}$$

$$f'(x)=x^{x^2}(2x^2\ln x + x^2+1)$$

Solution

We do not have a derivative formula for a function to the power of a function. To work around this, we use a “trick” – we use logarithm rules to get a multiplication of functions instead of a function to the power of a function.

$$f(x)=x^{x^2+1}$$

$$\ln f(x)=\ln x^{x^2+1}$$

$$\ln f(x)=(x^2+1)\ln x$$

Now, we divide both sides by the variable x. In the right side, we use the multiplication rule in Derivative Rules:

$$\frac{1}{f(x)}f'(x)=2x\cdot \ln x + (x^2+1)\frac{1}{x}$$

We isolate the derivative on one side:

$$f'(x)=f(x)(2x\cdot \ln x + (x^2+1)\frac{1}{x})$$

And set the function:

$$f'(x)=x^{x^2+1}(2x\ln x + (x^2+1)\frac{1}{x})$$

One can simplify the derivative:

$$f'(x)=x^{x^2}\cdot x(2x\ln x + (x^2+1)\frac{1}{x})$$

$$f'(x)=x^{x^2}\cdot x(2x\ln x + x+\frac{1}{x})$$

$$f'(x)=x^{x^2}(2x^2\ln x + x^2+1)$$

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