[tex]\frac{d(lnx)}{dx}=\frac{1}{x}[/tex]
Answer:
[tex]\frac{dy}{dx}=e^{x^2}(\frac{2x^2ln(4x^3)+3}{x})[/tex]
Step-by-step explanation:
We are given that a function
[tex]y=e^{x^2}ln(4x^3)[/tex]
We have to differentiate w.r.t x
[tex]\frac{dy}{dx}=e^{x^2}\times 2xln(4x^3)+e^{x^2}\times \frac{1}{4x^3}\times 12x^2[/tex]
By using formula
[tex]\frac{d(lnx)}{dx}=\frac{1}{x}[/tex]
[tex]\frac{d(e^x)}{dx}=e^x[/tex]
[tex]\frac{dy}{dx}=e^{x^2}(2xln(4x^3)+\frac{3}{x})[/tex]
[tex]\frac{dy}{dx}=e^{x^2}(\frac{2x^2ln(4x^3)+3}{x})[/tex]
Hence, the derivative of function
[tex]\frac{dy}{dx}=e^{x^2}(\frac{2x^2ln(4x^3)+3}{x})[/tex]
