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∫x3ex4 dx = 1/4ex4 + c To solve, let y = x4, then: dy = 4x3 dx ⇒ 1/4dy = x3 dx ⇒ ∫x3ex4 dx =∫ex4 x3 dx = ∫ey 1/4 dy = 1/4ey + c but y = x4, thus: = 1/4ex4 + c
The slope is dy/dx = 8/4 = 2
start by setting y=lnx^lnx take ln of both sides lny=lnx(ln(lnx)) differentiate dy/dx(1/y)=(1+ln(lnx))/x dy/dx=y(1+ln(lnx))/x we know that y=lnx^lnx so we can just substatute back in dy/dx=(lnx^lnx)*(1+ln(lnx))/x
d/dx (2cos(x)sin⁻¹(x)) right, 2cos(x)sin⁻¹(x) is a product, so we'll differentiate using the product rule: d/dx(uv) = u d/dx(v) + v d/dx(u) u = 2cos(x) v = sin⁻¹(x) d/dx(u) = -2sin(x) to find d/dx(sin⁻¹(x)) we'll set y=sin⁻¹(x) x=sin(y) dx/dy = cos(y) dy/dx = 1/(cos(y)) cos²(y) + sin²(y)=1, dy/dx = 1/sqrt(1-sin²y) = 1/sqrt(1-x²) [x=sin(y)] so plugging all this into our product rule, d/dx (2cos(x)sin⁻¹(x)) = 2cos(x)/sqrt(1-x²) - 2sin(x)sin⁻¹(x).
y = (sinx)^(e^x) ln(y) = ln((sinx)^(e^x)) ln(y) = (e^x)ln(sinx) (1/y)dy = (e^x)(1/sinx)(cosx)+ln(sinx)(e^x)dx (1/y)dy = (e^x)(cotx)+ln(sinx)(e^x)dx dy = ((sinx)^(e^x))((cotx)(e^x)+ln(sinx)(e^x))dx dy = ((e^x)(sinx)^(e^x))(cotx+ln(sinx))dx