How do I find the derivative of $y = log (x^{2} + 1)$ $y = log (x^2 + 1)$ ?

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How do I find the derivative of $y = log (x^{2} + 1)$ $y = log (x^2 + 1)$ ?

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Answer 1 · 2017-06-08T13:22:07

$\frac{d y}{d x} = \frac{2 x log e}{x^{2} + 1}$ $dy/dx=(2xlog e)/(x^2+1)$

Explanation:

The derivative of $y = log (u)$ $y=log(u)$ is $\frac{d y}{d x} = log e \cdot \frac{1}{u} \times \frac{d}{d x} (u)$ $dy/dx=log e cdot 1/u xx d/dx(u)$ , since $ln x = \frac{1}{log e} log x$ $lnx = 1/(log e) log x$ , and $\frac{d (ln x)}{d x} = \frac{1}{x}$ $(d(lnx))/(dx) = 1/x$ .

Following these exact steps we get:

$\frac{d y}{d x} = log e \cdot \frac{1}{x^{2} + 1} \times 2 x$ $dy/dx=log e cdot (1)/(x^2+1) xx 2x$

The final answer is:

$\frac{d y}{d x} = \frac{2 x log e}{x^{2} + 1}$ $dy/dx=(2xlog e)/(x^2+1)$

Answer 2 · 2017-06-09T16:58:55

$\frac{d y}{d x} = \frac{2 x}{(x^{2} + 1) ln 10}$ $dy/dx=(2x)/((x^2+1)ln10)$

Explanation:

$y = log (x^{2} + 1)$ $y=log(x^2+1)$

We can't differentiate $log (x^{2} + 1)$ $log(x^2+1)$ , so we must rewrite in such a way that we can. As such, the change of base rule will be required:

${log}_{a} b \equiv \frac{{log}_{c} b}{{log}_{c} a}$ $log_a b-=log_c b/log_c a$

$log (x^{2} + 1) \equiv \frac{ln (x^{2} + 1)}{ln 10}$ $log(x^2+1)-=ln(x^2+1)/ln10$

$y = \frac{ln (x^{2} + 1)}{ln 10}$ $y=ln(x^2+1)/ln10$

$\frac{d y}{d x} = \frac{1}{ln 10}$ $dy/dx=1/ln10$ $[\frac{d}{d x} (ln (x^{2} + 1))]$ $[d/dx(ln(x^2+1))]$

$\frac{d}{d x} (ln (x^{2} + 1)) = \frac{\frac{d}{d x} (x^{2} + 1)}{x^{2} + 1} = \frac{2 x}{x^{2} + 1}$ $d/dx(ln(x^2+1))=(d/dx(x^2+1))/(x^2+1)=(2x)/(x^2+1)$

$\frac{d y}{d x} = \frac{1}{ln 10} \times \frac{2 x}{x^{2} + 1} = \frac{2 x}{(x^{2} + 1) ln 10}$ $dy/dx=1/ln10xx(2x)/(x^2+1)=(2x)/((x^2+1)ln10)$

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How do I find the derivative of $y = log (x^{2} + 1)$ $y = log (x^2 + 1)$ ?

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How do I find the derivative of y=log(x2+1)y = log (x^2 + 1)?

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How do I find the derivative of $y = log (x^{2} + 1)$ $y = log (x^2 + 1)$ ?