tentukan dy/dx dari a.) y= e cos2x b.) y= X x atau kalo gak ngerti liat fotonya

Turunan.
Kalkulus tingkat perguruan tinggi.

Jawaban soal a
y = e^(cos² x)
dy / dx = e^(cos² x) [2 cos x (-sin x)]
            = -(sin 2x) e^(cos²x)

Jawaban soal b
y = x^x = e^(x ㏑ x)
dy / dx = e^(x ㏑ x) (1 ㏑ x + x 1/x)
            = x^x (㏑ x + 1)

[tex]\displaystyle y=e^{\cos^2x}\\\\\frac{dy}{dx}=(-2\cos x\sin x)e^{\cos^2x}\\\boxed{\boxed{\frac{dy}{dx}=-e^{\cos^2x}\sin2x}}\\\\y=x^x\\y=e^{\ln x^x}\\y=e^{x\ln x}\\\\\frac{dy}{dx}=e^{x\ln x}\cdot\left(\ln x+x\cdot\frac1x\right)\\\boxed{\boxed{\frac{dy}{dx}=x^x(\ln x+1)}}[/tex]