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Calculus Examples
y=3t(2t3-5)5y=3t(2t3−5)5
Step 1
Differentiate both sides of the equation.
ddt(y)=ddt(3t(2t3-5)5)ddt(y)=ddt(3t(2t3−5)5)
Step 2
The derivative of yy with respect to tt is y′.
y′
Step 3
Step 3.1
Since 3 is constant with respect to t, the derivative of 3t(2t3-5)5 with respect to t is 3ddt[t(2t3-5)5].
3ddt[t(2t3-5)5]
Step 3.2
Differentiate using the Product Rule which states that ddt[f(t)g(t)] is f(t)ddt[g(t)]+g(t)ddt[f(t)] where f(t)=t and g(t)=(2t3-5)5.
3(tddt[(2t3-5)5]+(2t3-5)5ddt[t])
Step 3.3
Differentiate using the chain rule, which states that ddt[f(g(t))] is f′(g(t))g′(t) where f(t)=t5 and g(t)=2t3-5.
Step 3.3.1
To apply the Chain Rule, set u as 2t3-5.
3(t(ddu[u5]ddt[2t3-5])+(2t3-5)5ddt[t])
Step 3.3.2
Differentiate using the Power Rule which states that ddu[un] is nun-1 where n=5.
3(t(5u4ddt[2t3-5])+(2t3-5)5ddt[t])
Step 3.3.3
Replace all occurrences of u with 2t3-5.
3(t(5(2t3-5)4ddt[2t3-5])+(2t3-5)5ddt[t])
3(t(5(2t3-5)4ddt[2t3-5])+(2t3-5)5ddt[t])
Step 3.4
Differentiate.
Step 3.4.1
By the Sum Rule, the derivative of 2t3-5 with respect to t is ddt[2t3]+ddt[-5].
3(t(5(2t3-5)4(ddt[2t3]+ddt[-5]))+(2t3-5)5ddt[t])
Step 3.4.2
Since 2 is constant with respect to t, the derivative of 2t3 with respect to t is 2ddt[t3].
3(t(5(2t3-5)4(2ddt[t3]+ddt[-5]))+(2t3-5)5ddt[t])
Step 3.4.3
Differentiate using the Power Rule which states that ddt[tn] is ntn-1 where n=3.
3(t(5(2t3-5)4(2(3t2)+ddt[-5]))+(2t3-5)5ddt[t])
Step 3.4.4
Multiply 3 by 2.
3(t(5(2t3-5)4(6t2+ddt[-5]))+(2t3-5)5ddt[t])
Step 3.4.5
Since -5 is constant with respect to t, the derivative of -5 with respect to t is 0.
3(t(5(2t3-5)4(6t2+0))+(2t3-5)5ddt[t])
Step 3.4.6
Simplify the expression.
Step 3.4.6.1
Add 6t2 and 0.
3(t(5(2t3-5)4(6t2))+(2t3-5)5ddt[t])
Step 3.4.6.2
Multiply 6 by 5.
3(t(30(2t3-5)4t2)+(2t3-5)5ddt[t])
3(t(30(2t3-5)4t2)+(2t3-5)5ddt[t])
3(t(30(2t3-5)4t2)+(2t3-5)5ddt[t])
Step 3.5
Raise t to the power of 1.
3(t2t1(30(2t3-5)4)+(2t3-5)5ddt[t])
Step 3.6
Use the power rule aman=am+n to combine exponents.
3(t2+1(30(2t3-5)4)+(2t3-5)5ddt[t])
Step 3.7
Add 2 and 1.
3(t3(30(2t3-5)4)+(2t3-5)5ddt[t])
Step 3.8
Differentiate using the Power Rule which states that ddt[tn] is ntn-1 where n=1.
3(t3(30(2t3-5)4)+(2t3-5)5⋅1)
Step 3.9
Multiply (2t3-5)5 by 1.
3(t3(30(2t3-5)4)+(2t3-5)5)
Step 3.10
Simplify.
Step 3.10.1
Apply the distributive property.
3(t3(30(2t3-5)4))+3(2t3-5)5
Step 3.10.2
Multiply 30 by 3.
90(t3((2t3-5)4))+3(2t3-5)5
Step 3.10.3
Factor 3(2t3-5)4 out of 90t3(2t3-5)4+3(2t3-5)5.
Step 3.10.3.1
Factor 3(2t3-5)4 out of 90t3(2t3-5)4.
3(2t3-5)4(30t3)+3(2t3-5)5
Step 3.10.3.2
Factor 3(2t3-5)4 out of 3(2t3-5)5.
3(2t3-5)4(30t3)+3(2t3-5)4(2t3-5)
Step 3.10.3.3
Factor 3(2t3-5)4 out of 3(2t3-5)4(30t3)+3(2t3-5)4(2t3-5).
3(2t3-5)4(30t3+2t3-5)
3(2t3-5)4(30t3+2t3-5)
Step 3.10.4
Add 30t3 and 2t3.
3(2t3-5)4(32t3-5)
3(2t3-5)4(32t3-5)
3(2t3-5)4(32t3-5)
Step 4
Reform the equation by setting the left side equal to the right side.
y′=3(2t3-5)4(32t3-5)
Step 5
Replace y′ with dydt.
dydt=3(2t3-5)4(32t3-5)