find $y'$ for $y=(4+x^2)^x$

Question

This differentiation requires the use of natural logarithms (the laws of logarithms), differentiation of logarithms, exponential function differentiation and the power rule.

the formula for differentiation of exponential functions is $d/dxa^x = a^x*ln(a)$

I use this to get $dy/dx = (4+x^2)^x*ln(4+x^2)$ but using a derivative calculator this is incorrect. Please help with where I go wrong.

Do I need to use this formula: $a^x = e^{xln(a)}$

Answer 1

$$y=(4+x^{ 2 })^{ x }\\ \ln { y } =x\ln { \left( 4+{ x }^{ 2 } \right) } \\ { \left( \ln { y } \right) }^{ \prime }={ \left( x\ln { \left( 4+{ x }^{ 2 } \right) } \right) }^{ \prime }\\ \frac { { y }^{ \prime } }{ y } =\ln { \left( 4+{ x }^{ 2 } \right) } +\frac { 2{ x }^{ 2 } }{ 4+{ x }^{ 2 } } \\ { y }^{ \prime }=(4+x^{ 2 })^{ x }\left[ \ln { \left( 4+{ x }^{ 2 } \right) } +\frac { 2{ x }^{ 2 } }{ 4+{ x }^{ 2 } } \right] \\ $$

Answer 2

HINT:

$$\frac{\text{d}}{\text{d}x}\left(\left(4+x^2\right)^x\right)=$$ $$\frac{\text{d}}{\text{d}x}\left(e^{x\ln\left(4+x^2\right)}\right)=$$

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Use the chain rule:

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$$e^{x\ln\left(4+x^2\right)}\cdot\frac{\text{d}}{\text{d}x}\left(x\ln\left(4+x^2\right)\right)=$$ $$(x^2+4)^x\cdot\frac{\text{d}}{\text{d}x}\left(x\ln\left(4+x^2\right)\right)=$$

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Use the product rule:

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$$(x^2+4)^x\cdot\left(x\left(\frac{\text{d}}{\text{d}x}\left(\ln\left(4+x^2\right)\right)\right)+1\cdot\ln(4+x^2)\right)=$$

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Use the chain rule:

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$$(4+x^2)^x\cdot\left(\ln(4+x^2)+x\cdot\frac{\frac{\text{d}}{\text{d}x}(4+x^2)}{x^2+4}\right)$$

Answer 3

It is better to derive the general derivative of the form $f(x)^x$ $$y=f(x)^x$$ $$\log y=x\log f(x)$$ take the derivative $$\frac{y'}{y}=1\log f(x)+x.\frac{f'(x)}{f(x)}$$ $$y'=y\log f(x)+yx.\frac{f'(x)}{f(x)}$$ $$y'=f(x)^x\log f(x)+f(x)^xx.\frac{f'(x)}{f(x)}$$ if the $f(x)=constant=a$ $$y'=a^x\log a+ax.\frac{0}{a}=a^x\log a$$

Answer 4

$$\begin{align} y&=(4+x^2)^x\\ \therefore \ln y&=x \cdot \ln (4+x^2)\\ \\ \text{differentiating w.r.t x}\\ \\ \frac{1}{y}\cdot \frac{dy}{dx}&= \ln (4+x^2)+x\cdot\frac{2x}{4+x^2}\\ \\ \frac{dy}{dx}= y\left(\ln (4+x^2)+x\cdot\frac{2x}{4+x^2}\right)&=\color{red}{\left(\left(4+x^2\right)^x\right)\left(\ln (4+x^2)+x\cdot\frac{2x}{4+x^2}\right)} \end{align}$$ Simplify the expression in red to get your answer.