How can i derive this function??

Question

I need to find $\frac{\mathrm{d}F }{\mathrm{d} x}$ for $F\left ( x \right )=\cosh^{\sin 2x } \left ( 6x \right )$

But i have no idea how to do it as every single one of my attempts has been wrong.

Answer 1

\begin{align} F(x)&=\cosh^{\sin 2x}(6x)\\ &=e^{\sin2x\cdot\ln[\cosh(6x)]}\\ \implies\dfrac{dF(x)}{dx}&=e^{\sin2x\cdot\ln[\cosh(6x)]}\cdot\dfrac{d}{dx}\left[\sin2x\cdot\ln[\cosh(6x)]\right]\\ &=\cosh^{\sin 2x}(6x)\cdot\left[\sin2x\cdot\dfrac{6\sinh(6x)}{\cosh(6x)}+\ln[\cosh(6x)]\cdot2\cos2x\right]\\ &=\cosh^{\sin 2x}(6x)\cdot\left[6\sin2x\cdot\tanh(6x)+\ln[\cosh(6x)]\cdot2\cos2x\right]\\ \end{align}

Answer 2

For this kind of problems, logarithmic differentiation is your friend $$F=\cosh^{\sin (2x) } \left ( 6x \right )$$ $$\log(F)=\sin(2x)\log(\cosh(6x))$$ Now, differentiate both sides $$\frac{F'}F=\left(\sin(2x)\right)'\log(\cosh(6x))+\sin(2x)\left(\log(\cosh(6x))\right)'$$ When finished, use $$F'=F \times \frac{F'}F$$