If y=2sin(3x)+e4xy = 2\sin(3x) + e^{4x}y=2sin(3x)+e4x, what is the value of dydx\frac{dy}{dx}dxdy at x=0x = 0x=0? Cevap: 10Cevap10Differentiating y=2sin(3x)+e4xy = 2\sin(3x) + e^{4x}y=2sin(3x)+e4x yields dydx=6cos(3x)+4e4x\frac{dy}{dx} = 6\cos(3x) + 4e^{4x}dxdy=6cos(3x)+4e4x. Evaluating this expression at x=0x = 0x=0 gives 6cos(0)+4e0=6(1)+4(1)=106\cos(0) + 4e^{0} = 6(1) + 4(1) = 106cos(0)+4e0=6(1)+4(1)=10.Adım Adım Çözüm1Differentiate each term of y=2sin(3x)+e4xy = 2\sin(3x) + e^{4x}y=2sin(3x)+e4x with respect to xxx.dydx=6cos(3x)+4e4x\frac{dy}{dx} = 6\cos(3x) + 4e^{4x}dxdy=6cos(3x)+4e4xApplying the chain rule gives ddx[2sin(3x)]=2×3cos(3x)=6cos(3x)\frac{d}{dx}[2\sin(3x)] = 2 \times 3\cos(3x) = 6\cos(3x)dxd[2sin(3x)]=2×3cos(3x)=6cos(3x) and ddx[e4x]=4e4x\frac{d}{dx}[e^{4x}] = 4e^{4x}dxd[e4x]=4e4x.2Evaluate the derivative at x=0x = 0x=0.6\cos(0) + 4e^{0} = 6(1) + 4(1) = 10Substituting x=0x = 0x=0 gives cos(0)=1\cos(0) = 1cos(0)=1 and e0=1e^{0} = 1e0=1.Anahtar KavramDifferentiation of trigonometric and exponential functionsSık Yapılan Hatalar