Question

differentiate 5^x logx / x^2 + 1​

Answer 1

$\underline{\textbf{Given:}}$

$\mathsf{\dfrac{5^x\;logx}{x^2+1}}$

$\underline{\textbf{To find:}}$

$\textsf{Derivative of the given function}$

$\underline{\textbf{Solution:}}$

$\underline{\textsf{Concept used:}}$

$\mathsf{Product\;rule:\;\dfrac{d(uv)}{dx}=u\;\dfrac{dv}{dx}+v\;\dfrac{du}{dx}}$

$\mathsf{Quotient\;rule:\;\dfrac{d(\frac{u}{v})}{dx}=\dfrac{v\,\dfrac{du}{dx}-u\,\dfrac{dv}{dx}}{v^2}}$

$\mathsf{Take,\;\;y=\dfrac{5^x\;logx}{x^2+1}}$

$\textsf{Differentiate with respect to 'x'}$

$\mathsf{\dfrac{dy}{dx}=\dfrac{(x^2+1)\dfrac{d(5^x\,logx)}{dx}-(5^x\,logx)\dfrac{d(x^2+1)}{dx}}{(x^2+1)^2}}$

$\mathsf{\dfrac{dy}{dx}=\dfrac{(x^2+1)\left[5^x\;\left(\dfrac{1}{x}\right)+logx\;5^x\,log5\right]-(5^x\,logx)2x}{(x^2+1)^2}}$

$\mathsf{\dfrac{dy}{dx}=\dfrac{(x^2+1)\left[\dfrac{5^x}{x}+logx\;5^x\,log5\right]-(5^x\,logx)2x}{(x^2+1)^2}}$

$\mathsf{\dfrac{dy}{dx}=\dfrac{(x^2+1)\dfrac{5^x}{x}+(x^2+1)logx\;5^x\,log5-(5^x\,logx)2x}{(x^2+1)^2}}$

$\therefoer\boxed{\mathsf{\dfrac{dy}{dx}=\dfrac{(x^2+1)\dfrac{5^x}{x}+5^x\,logx\left[(x^2+1)\,log5-2x\right]}{(x^2+1)^2}}}$

$\underline{\textbf{Find more:}}$

If y = tan-1(6x-7/6+7x)

then dy/dx is

​https://brainly.in/question/36894600  

y=tan^-1 1+x²/1-x² then find dy/dx=?  

https://brainly.in/question/12313077

Answer 2

Step-by-step explanation:

•in step 1 we differentiate by product rule and quotient rule

•in step 2 we just solve it

•in step 3 we take the LCM of X and put the in denominator by reverse

•in step 4 common 5^x ,