Question

find the derivative of (cosx)^x+ x^cosx.? plz help me fastly..
${cosx}^{x} + {x}^{cosx}$
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Answer 1

Step-by-step explanation:

Consider the general formula

Let $y={f(x)}^{g(x)}$

Then,

$\ln(y) = g(x). \ln \{ f(x)\}$

$\implies \frac{1}{y} . \frac{dy}{dx} = \frac{d}{dx} \bigg[ g(x). \ln \{ f(x)\} \bigg]$

$\implies \frac{dy}{dx} =y. \frac{d}{dx} \bigg[ g(x). \ln \{ f(x)\} \bigg]$

$\implies \frac{dy}{dx} = { \{ f(x)\}}^{g(x)} . \frac{d}{dx} \bigg[ g(x). \ln \{ f(x)\} \bigg]$

Using this formula, we will find the derivative of the given question

Now,

Let,

$y = \{ \cos(x) \}^{x} + { \{x \}}^{ \cos(x) }$

$\implies \: \frac{dy}{dx} = \{ \cos(x) \}^{x} . \frac{d}{dx} \bigg \{ x. \ln( \cos(x) ) \bigg\} + { \{x \}}^{ \cos(x) } . \frac{d}{dx} \bigg \{ \cos(x) . \ln(x) \bigg\}$

$\implies \: \frac{dy}{dx} = \{ \cos(x) \}^{x} . \bigg \{ \ln( \cos(x) )\frac{d}{dx} (x) + x\frac{d}{dx} \{ \ln( \cos(x) ) \}\bigg\} + { \{x \}}^{ \cos(x) } . \bigg \{ \ln(x) \frac{d}{dx} \{\cos(x) \} + \cos(x) \frac{d}{dx} \{ \ln(x) \} \bigg\}$

$\implies \: \frac{dy}{dx} = \{ \cos(x) \}^{x} . \bigg \{ \ln( \cos(x) ).1 + x. \frac{ - \sin(x) }{ \cos(x) } \bigg\} + { \{x \}}^{ \cos(x) } . \bigg \{ \ln(x) . - \sin(x) + \cos(x). \frac{1}{x} \bigg\}$

$\implies \: \frac{dy}{dx} = \{ \cos(x) \}^{x} . \bigg \{ \ln( \cos(x) ) - x \tan(x) \bigg\} + { \{x \}}^{ \cos(x) } . \bigg \{ - \sin(x). \ln(x) + \frac{ \cos(x) }{x} \bigg\}$