I know I'll have to use implicit differentiation, but I always get stuck when there is an exponent with trig, log, and/or natural log.
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7As currently written, the expression is difficult to decipher. – André Nicolas Nov 15 '14 at 05:22
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@Kaaagome pls check edit – Sophie Clad Nov 15 '14 at 06:50
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Approach 1$$y=x^{y^{\sin x}}$$ $$\ln y=y^{\sin x}\ln x$$ $$\frac1y\frac{dy}{dx}=\frac{y^{\sin x}}{x}+\ln x\left[\frac{d}{dx}y^{\sin x}\right]$$
Now find derivative of $y^{\sin x}$ and then rearrngement
Approach 2 $$y=x^{y^{\sin x}}$$
$$\ln y=y^{\sin x}\ln x$$ $$\ln(\ln y)=\ln(y^{\sin x})+\ln(\ln x)$$
$$\ln(\ln y)={\sin x}\ln(y)+\ln(\ln x)$$
I hope You can take it from here!
Note: Question was edited after posting answer given below
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$$((x^{y})^{sinx})^{6}=x^{6y\sin x}$$
$$y=x^{6y\sin x}$$ $$\ln y=6y\sin x\ln x$$
$$\frac1y\frac{dy}{dx}=6\sin x\ln x\frac{dy}{dx}+\frac{6y\sin x}{x}+6y\cos x\ln x$$ $$\left(\frac1y-6\sin x\ln x\right)\frac{dy}{dx}=\frac{6y\sin x}{x}+6y\cos x\ln x$$ $$\left(\frac{1-6y\sin x\ln x}{y}\right)\frac{dy}{dx}=\frac{6y\sin x}{x}+6y\cos x\ln x$$
$$\frac{dy}{dx}=\frac{y(6y\sin x+6xy\cos x\ln x)}{x(1-6y\sin x\ln x)}$$
Now it's all about simplification, Just like last time
Aditya Hase
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thank you for the answer, but someone edited the title wrong. it's y=x^(y^sinx) – Elsa Nov 16 '14 at 06:23
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Then $\dfrac{dy}{dx}=\dfrac{y(y\sin x+xy\cos x\ln x)}{x(1-y\sin x\ln x)}$ – Teoc Nov 16 '14 at 06:28
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oh my gosh thank you so much! approach 2 makes alot of sense! i tried something like that, but i forgot that property where u can add the two ln's together! thank you! – Elsa Nov 16 '14 at 06:50